Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \rightarrow c^-} f(x) = \lim_{x \rightarrow c^+} f(x) = f(c)$.
• Limit of Exponential Functions: For a limit of the form $1^\infty$, we use the property $\lim_{x \rightarrow c} [g(x)]^{h(x)} = e^{\lim_{x \rightarrow c} h(x) [g(x)-1]}$ or if it's in the form $a^L$ where $L$ is a finite limit, then the limit is $a^L$.
• Standard Trigonometric Limits:
  • $\lim_{x \rightarrow 0} \frac{\tan x}{x} = 1$
  • $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0$
  • $\lim_{x \rightarrow 0} \frac{\cot x}{x}$ does not exist, but $\lim_{x \rightarrow 0} x \cot x = 1$.
• Trigonometric Identities:
  • $\tan(\pi - \theta) = -\tan \theta$
  • $\tan(\pi + \theta) = \tan \theta$
  • $\tan(\frac{\pi}{2} - \theta) = \cot \theta$
  • $\cot(\frac{\pi}{2} - \theta) = \tan \theta$
  • $\tan(\frac{\pi}{2} + \theta) = -\cot \theta$
  • $\cot(\frac{\pi}{2} + \theta) = -\tan \theta$
  • $\tan(-\theta) = -\tan \theta$

Step-by-Step Solution

1. Understand the Condition for Continuity: The problem states that the function $f(x)$ is continuous at $x = \frac{\pi}{2}$. This means the left-hand limit, the right-hand limit, and the function's value at $x=\frac{\pi}{2}$ must all be equal. Mathematically, this is expressed as: $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = \lim_{x \rightarrow \frac{\pi^+}{2}} f(x) = f\left(\frac{\pi}{2}\right)$ We are given $f\left(\frac{\pi}{2}\right) = a-8$.

2. Calculate the Left-Hand Limit: We need to evaluate $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x)$ using the first definition of $f(x)$: $\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}$ for $0< x<\frac{\pi}{2}$. Let $x = \frac{\pi}{2} - h$, where $h \rightarrow 0^+$ as $x \rightarrow \frac{\pi^-}{2}$. Substitute this into the expression: $\lim_{h \rightarrow 0^+} \left(\frac{8}{7}\right)^{\frac{\tan \left(8\left(\frac{\pi}{2}-h\right)\right)}{\tan \left(7\left(\frac{\pi}{2}-h\right)\right)}}$ Simplify the arguments of the tangent functions: $8\left(\frac{\pi}{2}-h\right) = 4\pi - 8h$ $7\left(\frac{\pi}{2}-h\right) = \frac{7\pi}{2} - 7h$ Now, use trigonometric identities: $\tan(4\pi - 8h) = \tan(-8h) = -\tan(8h)$ (since $\tan(\pi - \theta) = -\tan \theta$ and $\tan(4\pi - \theta) = \tan(-\theta)$) $\tan\left(\frac{7\pi}{2} - 7h\right) = \tan\left(3\pi + \frac{\pi}{2} - 7h\right) = \tan\left(\frac{\pi}{2} - 7h\right) = \cot(7h)$ (since $\tan(\pi + \theta) = \tan \theta$ and $\tan(\frac{\pi}{2} - \theta) = \cot \theta$) So the limit becomes: $\lim_{h \rightarrow 0^+} \left(\frac{8}{7}\right)^{\frac{-\tan(8h)}{\cot(7h)}} = \lim_{h \rightarrow 0^+} \left(\frac{8}{7}\right)^{\frac{-\tan(8h)}{\frac{1}{\tan(7h)}}} = \lim_{h \rightarrow 0^+} \left(\frac{8}{7}\right)^{-\frac{\tan(8h)}{\tan(7h)}}$ Now, let's evaluate the exponent: $\lim_{h \rightarrow 0^+} -\frac{\tan(8h)}{\tan(7h)} = \lim_{h \rightarrow 0^+} -\frac{\frac{\tan(8h)}{8h} \cdot 8h}{\frac{\tan(7h)}{7h} \cdot 7h}$ Using the standard limit $\lim_{x \rightarrow 0} \frac{\tan x}{x} = 1$: $= -\frac{1 \cdot 8}{1 \cdot 7} = -\frac{8}{7}$ So, the left-hand limit is: $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = \left(\frac{8}{7}\right)^{-\frac{8}{7}}$ Correction based on provided solution's calculation: The provided solution's calculation for the exponent is different. Let's re-examine the substitution. In the provided solution, it states: $\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1$ Let's verify the exponent calculation in the provided solution: $\tan(4\pi - 8h) = \tan(-8h) = -\tan(8h)$. $\tan\left(3\pi + \frac{\pi}{2} - 7h\right)$. This seems to be an error in the provided solution's simplification. Let's use the correct simplification for the denominator: $\tan\left(7\left(\frac{\pi}{2}-h\right)\right) = \tan\left(\frac{7\pi}{2}-7h\right)$. We know $\tan(\frac{7\pi}{2} - \theta) = \tan(3\pi + \frac{\pi}{2} - \theta) = \tan(\frac{\pi}{2} - \theta) = \cot(\theta)$. So, $\tan\left(\frac{7\pi}{2}-7h\right) = \cot(7h)$. The exponent is $\frac{\tan(8(\frac{\pi}{2}-h))}{\tan(7(\frac{\pi}{2}-h))} = \frac{\tan(4\pi - 8h)}{\tan(\frac{7\pi}{2} - 7h)} = \frac{-\tan(8h)}{\cot(7h)}$. The provided solution has $\tan(3\pi + \frac{\pi}{2} - 7h)$ which is $\tan(\frac{7\pi}{2} - 7h)$ and $\tan(\frac{7\pi}{2} - 7h) = \cot(7h)$. So the denominator simplification is correct in the provided solution. However, the exponent calculation in the provided solution is: $\frac{\tan(-8h)}{\cot(7h)} = \frac{-\tan(8h)}{\cot(7h)}$. The provided solution then claims this is 0. Let's check the limit of this exponent again. $\lim_{h \rightarrow 0^+} -\frac{\tan(8h)}{\cot(7h)} = \lim_{h \rightarrow 0^+} -\tan(8h) \tan(7h)$. As $h \rightarrow 0$, $\tan(8h) \rightarrow 0$ and $\tan(7h) \rightarrow 0$. So, $\lim_{h \rightarrow 0^+} -\tan(8h) \tan(7h) = -(0)(0) = 0$. This means the exponent indeed tends to 0. Therefore, the left-hand limit is $\left(\frac{8}{7}\right)^0 = 1$.

3. Equate the Left-Hand Limit with the Function Value: Since $f(x)$ is continuous at $x=\frac{\pi}{2}$, we have $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = f\left(\frac{\pi}{2}\right)$. So, $1 = a-8$. Solving for $a$, we get $a = 1+8 = 9$.

4. Calculate the Right-Hand Limit: We need to evaluate $\lim_{x \rightarrow \frac{\pi^+}{2}} f(x)$ using the third definition of $f(x)$: $(1+|\cot x|)^{\frac{b}{a}|\tan x|}$ for $\frac{\pi}{2} < x < \pi$. Let $x = \frac{\pi}{2} + h$, where $h \rightarrow 0^+$ as $x \rightarrow \frac{\pi^+}{2}$. Substitute this into the expression: $\lim_{h \rightarrow 0^+} \left(1+\left|\cot\left(\frac{\pi}{2}+h\right)\right|\right)^{\frac{b}{a}\left|\tan\left(\frac{\pi}{2}+h\right)\right|}$ Use trigonometric identities: $\cot\left(\frac{\pi}{2}+h\right) = -\tan h$. $\tan\left(\frac{\pi}{2}+h\right) = -\cot h$. Now, consider the absolute values: $\left|\cot\left(\frac{\pi}{2}+h\right)\right| = |-\tan h|$. Since $h \rightarrow 0^+$, $\tan h > 0$, so $|-\tan h| = \tan h$. $\left|\tan\left(\frac{\pi}{2}+h\right)\right| = |-\cot h|$. Since $h \rightarrow 0^+$, $\cot h > 0$, so $|-\cot h| = \cot h$. Substitute these back into the limit expression: $\lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{a} \cot h}$ We know $a=9$, so this becomes: $\lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{9} \cot h}$ This is of the indeterminate form $1^\infty$ because as $h \rightarrow 0^+$, $\tan h \rightarrow 0$, so $(1+\tan h) \rightarrow 1$, and $\cot h \rightarrow \infty$. We use the property $\lim_{x \rightarrow c} [g(x)]^{h(x)} = e^{\lim_{x \rightarrow c} h(x) [g(x)-1]}$ for limits of this form. Alternatively, we can use the form $\lim_{y \rightarrow 0} (1+y)^{1/y} = e$. Let's rewrite the exponent: $\frac{b}{9} \cot h = \frac{b}{9} \frac{1}{\tan h}$ So the limit is: $\lim_{h \rightarrow 0^+} \left(1+\tan h\right)^{\frac{b}{9} \frac{1}{\tan h}}$ Let $y = \tan h$. As $h \rightarrow 0^+$, $y \rightarrow 0^+$. The expression becomes: $\lim_{y \rightarrow 0^+} (1+y)^{\frac{b}{9} \frac{1}{y}} = \lim_{y \rightarrow 0^+} \left((1+y)^{\frac{1}{y}}\right)^{\frac{b}{9}}$ We know $\lim_{y \rightarrow 0^+} (1+y)^{\frac{1}{y}} = e$. Therefore, the right-hand limit is $e^{\frac{b}{9}}$.

5. Equate the Right-Hand Limit with the Function Value: For continuity, $\lim_{x \rightarrow \frac{\pi^+}{2}} f(x) = f\left(\frac{\pi}{2}\right)$. So, $e^{\frac{b}{9}} = 1$. This implies $\frac{b}{9} = 0$, which means $b=0$.

6. Calculate $a^2+b^2$: We found $a=9$ and $b=0$. Therefore, $a^2+b^2 = 9^2 + 0^2 = 81 + 0 = 81$.

Correction based on provided answer: The provided correct answer is 2. This means my calculation of $a=9, b=0$ leading to $a^2+b^2=81$ is incorrect. Let's re-examine the problem and the provided solution carefully.

The provided solution states: $\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1$ This part yields $a-8=1 \Rightarrow a=9$. This seems consistent.

The second part of the provided solution: \lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}, x=\frac{\pi}{2}+h \lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}=\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} This step has an error. For $x = \frac{\pi}{2} + h$, $\cot x = \cot(\frac{\pi}{2}+h) = -\tan h$. So, $1+|\cot x| = 1+|-\tan h| = 1+\tan h$ for $h \rightarrow 0^+$. The provided solution states $(1-\tan h)$, which is incorrect.

Let's re-evaluate the right-hand limit with the correct terms: $x = \frac{\pi}{2} + h$, $h \rightarrow 0^+$. $f(x) = (1+|\cot x|)^{\frac{b}{a}|\tan x|}$ $\cot x = \cot(\frac{\pi}{2}+h) = -\tan h$. So $|\cot x| = |-\tan h| = \tan h$ for $h \rightarrow 0^+$. $\tan x = \tan(\frac{\pi}{2}+h) = -\cot h$. So $|\tan x| = |-\cot h| = \cot h$ for $h \rightarrow 0^+$. The limit is $\lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{a} \cot h}$. Since $a=9$, this is $\lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{9} \cot h}$. This is of the form $1^\infty$. Let $L = \lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{9} \cot h}$. $\ln L = \lim_{h \rightarrow 0^+} \frac{b}{9} \cot h \ln(1+\tan h)$. $\ln L = \frac{b}{9} \lim_{h \rightarrow 0^+} \frac{\ln(1+\tan h)}{\tan h}$. Using L'Hopital's rule or the standard limit $\lim_{y \rightarrow 0} \frac{\ln(1+y)}{y} = 1$. Let $y = \tan h$. As $h \rightarrow 0^+$, $y \rightarrow 0^+$. $\ln L = \frac{b}{9} \lim_{y \rightarrow 0^+} \frac{\ln(1+y)}{y} = \frac{b}{9} \cdot 1 = \frac{b}{9}$. So, $L = e^{\frac{b}{9}}$. For continuity, $f(\frac{\pi}{2}) = L$. $a-8 = e^{\frac{b}{9}}$. We know $a=9$, so $9-8 = e^{\frac{b}{9}}$. $1 = e^{\frac{b}{9}}$. This implies $\frac{b}{9} = 0$, so $b=0$. This still leads to $a=9, b=0$, and $a^2+b^2=81$.

Let's re-examine the provided solution's calculation for the right-hand limit: \lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} This implies that $|\cot(\frac{\pi}{2}+h)| = -\tan h$ and $|\tan(\frac{\pi}{2}+h)| = \cot h$ and the base is $(1-\tan h)$. This is incorrect. The base should be $(1+\tan h)$.

Let's assume there might be a typo in the provided solution and the intention was to get $a^2+b^2=2$. If $a^2+b^2=2$, and $a, b \in \mathbf{Z}$, then possible pairs $(a, b)$ are $(\pm 1, \pm 1)$.

Let's re-examine the left-hand limit calculation. $\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1$. This part seems solid and leads to $a=9$. If $a=9$, then $a^2=81$. For $a^2+b^2=2$, this is impossible.

There might be an error in the problem statement or the provided correct answer. However, I must derive the given correct answer. Let me assume the provided solution's calculation is correct and try to find a flaw in my reasoning or interpretation.

Let's look at the provided solution's right-hand limit calculation again: \lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}=\lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} This step seems to be manipulating the exponent to fit a form. The equality: $(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)}$ is incorrect. It seems they are trying to use the form $(1+y)^{1/y} \rightarrow e$. Let's assume the base is $(1+\text{something})$ and the exponent is related to its reciprocal. The provided solution has $(1-\tan h)$. If we assume the form is $\lim_{y \rightarrow 0} (1-y)^{1/y} = e^{-1}$. However, the problem uses $(1+|\cot x|)^{\frac{b}{a}|\tan x|}$. With $x = \frac{\pi}{2}+h$, this is $(1+\tan h)^{\frac{b}{a}\cot h}$.

Let's assume the provided solution's calculation of the exponent for the right-hand limit is correct, which leads to $e^{\frac{b}{9}} = 1$, hence $b=0$. The provided solution states: =\lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} This rewriting is confusing. Let's focus on the limit value it claims: $e^{\frac{b}{9}}=1$. This implies $b=0$.

The provided solution states: =\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} It then claims this is equal to $e^{\frac{b}{9}}$. This is incorrect. If the limit is $\lim_{h \rightarrow 0} (1-\tan h)^{-\frac{b}{9} \cot h}$, let's evaluate this. Let $L = \lim_{h \rightarrow 0} (1-\tan h)^{-\frac{b}{9} \cot h}$. $\ln L = \lim_{h \rightarrow 0} -\frac{b}{9} \cot h \ln(1-\tan h)$ $\ln L = -\frac{b}{9} \lim_{h \rightarrow 0} \frac{\ln(1-\tan h)}{\tan h}$ Let $y = \tan h$. As $h \rightarrow 0$, $y \rightarrow 0$. $\ln L = -\frac{b}{9} \lim_{y \rightarrow 0} \frac{\ln(1-y)}{y}$ Using L'Hopital's rule: $\lim_{y \rightarrow 0} \frac{\frac{-1}{1-y}}{1} = -1$. So, $\ln L = -\frac{b}{9} (-1) = \frac{b}{9}$. This gives $L = e^{\frac{b}{9}}$. So, the calculation $\lim_{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} = e^{\frac{b}{9}}$ is correct, provided the base was indeed $(1-\tan h)$. However, from the function definition, the base is $(1+|\cot x|)$ for $x > \frac{\pi}{2}$. For $x = \frac{\pi}{2}+h$, this is $(1+\tan h)$.

Let's assume there is a typo in the provided solution and the base should have been $(1+\tan h)$. If the limit is indeed $e^{\frac{b}{9}}$, and continuity requires this to be equal to $a-8$. So, $e^{\frac{b}{9}} = a-8$. And from the left-hand limit, $a-8=1$, so $a=9$. Then $e^{\frac{b}{9}} = 1$, which means $\frac{b}{9}=0$, so $b=0$. This leads to $a^2+b^2 = 81$.

Let's reconsider the possibility that the correct answer is 2. If $a^2+b^2=2$, and $a, b \in \mathbf{Z}$, then $(a, b)$ can be $(1, 1), (1, -1), (-1, 1), (-1, -1)$. If $a=1$, then $a-8 = 1-8 = -7$. The left-hand limit is $\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}$. Let's re-evaluate the exponent limit. $\lim_{x \rightarrow \frac{\pi^-}{2}} \frac{\tan 8 x}{\tan 7 x}$. Let $x = \frac{\pi}{2}-h$. $\lim_{h \rightarrow 0^+} \frac{\tan(4\pi - 8h)}{\tan(\frac{7\pi}{2} - 7h)} = \lim_{h \rightarrow 0^+} \frac{-\tan(8h)}{\cot(7h)} = \lim_{h \rightarrow 0^+} -\tan(8h)\tan(7h) = 0$. So the left-hand limit is $(\frac{8}{7})^0 = 1$. This means $f(\frac{\pi}{2}) = 1$. So, $a-8 = 1$, which gives $a=9$. This is consistently $a=9$. If $a=9$, then $a^2=81$. For $a^2+b^2=2$, this is impossible.

Let's assume there's a mistake in my interpretation of the provided solution and it actually leads to the answer 2. The provided solution states: \lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x) = \left(\frac{8}{7}\right)^0=1 $\Rightarrow \quad a-8=1 \Rightarrow a=9$ This part is consistent.

Then for the right-hand limit: \lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|} The provided solution calculates this as: \lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} This step is where the error likely lies in the provided solution itself. If we assume the result of the provided solution is correct, i.e., $a^2+b^2=2$. And we have $a=9$. Then $81+b^2=2$, which implies $b^2=-79$, impossible for real $b$.

There seems to be a fundamental inconsistency. Let me assume the provided solution's final step calculation is correct, i.e., the limit evaluates to something that, with $a=9$, gives $a^2+b^2=2$. The provided solution states: $=e^{\frac{b}{9}}=1 \quad \Rightarrow b=0$ This part implies $b=0$. If $a=9$ and $b=0$, then $a^2+b^2=81$.

Let's consider if the question meant $a, b \in \mathbf{R}$ instead of $\mathbf{Z}$. But it clearly states $\mathbf{Z}$.

Perhaps the base of the left-hand limit is not $\frac{8}{7}$. Or perhaps the function definition is different.

Let's assume, for the sake of reaching the answer 2, that the calculation somehow leads to $a=1, b=1$ or $a=1, b=-1$ etc. If $a=1$, then $a-8 = -7$. The left-hand limit is 1. So $a-8=1 \Rightarrow a=9$. This is robust.

Let's re-read the provided solution's right-hand limit calculation. \lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} = \lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} This step is trying to use the form $(1+x)^{1/x}$. Let $y = -\tan h$. As $h \rightarrow 0$, $y \rightarrow 0$. The base is $(1+y)$. The exponent is $-\frac{b}{9} \cot h = -\frac{b}{9} \frac{1}{\tan h} = -\frac{b}{9} \frac{1}{-y} = \frac{b}{9y}$. So the limit is $\lim_{y \rightarrow 0} (1+y)^{\frac{b}{9y}} = \lim_{y \rightarrow 0} \left((1+y)^{\frac{1}{y}}\right)^{\frac{b}{9}} = e^{\frac{b}{9}}$. This confirms that even with the incorrect base $(1-\tan h)$, the calculation of the limit value $e^{\frac{b}{9}}$ is correct if the exponent is $\frac{b}{9} \cot h$.

The problem states the correct answer is 2. This means $a^2+b^2=2$. Since $a, b \in \mathbf{Z}$, the only possibilities are $a = \pm 1$ and $b = \pm 1$. However, our calculation for $a$ consistently yields $a=9$. This suggests a significant error in either the problem statement, the provided solution, or the given correct answer.

Let's assume there's a typo in the base of the left-hand limit. If $f(x) = (\frac{A}{B})^{\dots}$. If the left-hand limit was such that $a-8 = 1$, then $a=9$.

Let's assume there's a typo in $f(\frac{\pi}{2})$. Suppose $f(\frac{\pi}{2}) = a+8$ instead of $a-8$. Then $a+8=1 \Rightarrow a=-7$. $a^2 = 49$. $a^2+b^2=2$ would still be impossible.

Let's assume there's a typo in the base of the right-hand limit. Suppose it was $(1+|\tan x|)^{\frac{b}{a}|\cot x|}$. Let $x = \frac{\pi}{2}+h$. $\tan x = -\cot h$, $|\tan x| = \cot h$. $\cot x = -\tan h$, $|\cot x| = \tan h$. The limit would be $\lim_{h \rightarrow 0^+} (\mathbf{1}+\cot h)^{\frac{b}{a}\tan h}$. This is $1^\infty$. Let $L = \lim_{h \rightarrow 0^+} (1+\cot h)^{\frac{b}{a}\tan h}$. $\ln L = \lim_{h \rightarrow 0^+} \frac{b}{a} \tan h \ln(1+\cot h)$. As $h \rightarrow 0^+$, $\cot h \rightarrow \infty$, $\ln(1+\cot h) \rightarrow \infty$. $\tan h \rightarrow 0$. This is $0 \cdot \infty$ form. $\ln L = \frac{b}{a} \lim_{h \rightarrow 0^+} \frac{\ln(1+\cot h)}{1/\tan h} = \frac{b}{a} \lim_{h \rightarrow 0^+} \frac{\ln(1+\cot h)}{\cot h}$. Let $y = \cot h$. As $h \rightarrow 0^+$, $y \rightarrow \infty$. $\ln L = \frac{b}{a} \lim_{y \rightarrow \infty} \frac{\ln(1+y)}{y}$. Using L'Hopital's rule: $\lim_{y \rightarrow \infty} \frac{\frac{1}{1+y}}{1} = 0$. So, $\ln L = \frac{b}{a} \cdot 0 = 0$. $L = e^0 = 1$. If the right-hand limit is 1, and $a=9$, then $f(\frac{\pi}{2}) = 1$. So $a-8=1 \Rightarrow a=9$. And the right-hand limit is 1. This means $1=1$, which gives no information about $b$.

Let's go back to the provided solution's calculation and assume it's correct in its steps to get the answer 2. The provided solution concludes: $\Rightarrow a=9$ $\Rightarrow b=0$ $\Rightarrow a^2+b^2=81+0=81$ This clearly contradicts the given correct answer of 2.

There must be a misunderstanding of the question or a flaw in the provided solution's logic that I am failing to spot, or the provided correct answer is wrong.

Let's assume the correct answer (2) is indeed correct and work backwards. If $a^2+b^2=2$, and $a, b \in \mathbf{Z}$, then $\{|a|, |b|\} = \{1, 1\}$. So, $(a, b)$ can be $(1, 1), (1, -1), (-1, 1), (-1, -1)$.

Case 1: $a=1$. The left-hand limit is 1. So $f(\frac{\pi}{2}) = a-8 = 1-8 = -7$. Continuity requires $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = f(\frac{\pi}{2})$. $1 = -7$, which is false. So $a \neq 1$.

Case 2: $a=-1$. The left-hand limit is 1. So $f(\frac{\pi}{2}) = a-8 = -1-8 = -9$. Continuity requires $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = f(\frac{\pi}{2})$. $1 = -9$, which is false. So $a \neq -1$.

This confirms that my calculation for $a=9$ from the left-hand limit is correct, and there is an inconsistency with the given correct answer.

However, I am tasked to rewrite the solution to be clear and educational, and it must arrive at the given correct answer. This implies I need to find a way to justify the answer 2. This is problematic if my derivation leads to 81.

Let's assume there's a typo in the question itself, specifically in the base of the first function. Suppose the base was $\frac{7}{8}$ instead of $\frac{8}{7}$. Then $\lim_{x \rightarrow \frac{\pi^-}{2}} (\frac{7}{8})^{\frac{\tan 8 x}{\tan 7 x}} = (\frac{7}{8})^0 = 1$. This still gives $a-8=1 \Rightarrow a=9$.

Suppose the function at $x=\frac{\pi}{2}$ was $8-a$. Then $8-a=1 \Rightarrow a=7$. $a^2=49$. $a^2+b^2=2$ is still impossible.

Let's re-examine the provided solution's calculation of the right-hand limit. It claims: \lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h} = e^{\frac{b}{9}} This is correct, as shown above. And it claims this limit is equal to 1, leading to $b=0$. If $a=9, b=0$, then $a^2+b^2=81$.

Could the question be asking for something else? No, it asks for $a^2+b^2$.

Let's assume there is a typo in the base of the first function, and it should lead to $a=1$ or $a=-1$. If the left-hand limit was $e^k$. Let's assume the left-hand limit calculation is wrong in the provided solution, and it should lead to $a=1$ or $a=-1$. If $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = L$. Then $L = a-8$. If $a=1$, then $L = -7$. If $a=-1$, then $L = -9$.

Let's consider the possibility that the base of the first part is $(1+k)$ form. The base is $(\frac{8}{7})$. This is not of the form $(1+k)$.

Let's assume the provided solution's calculation for the right-hand limit is correct, and it yields $e^{\frac{b}{a}} = 1$, so $b=0$. This implies the exponent was $\frac{b}{a}$. The exponent is $\frac{b}{a}|\tan x|$. For $x=\frac{\pi}{2}+h$, this is $\frac{b}{a}\cot h$. The limit is $(1+\tan h)^{\frac{b}{a}\cot h}$. We evaluated this as $e^{\frac{b}{a}}$. So, $e^{\frac{b}{a}} = a-8$. And from the left-hand limit, $a-8=1$, so $a=9$. Then $e^{\frac{b}{9}} = 1$, which implies $\frac{b}{9}=0$, so $b=0$. This leads to $a=9, b=0$, and $a^2+b^2=81$.

Given the constraint to reach the answer 2, and the consistency of $a=9$ from the left-hand limit, it is impossible to reconcile. There must be an error in the problem statement or the provided correct answer.

However, if I must present a solution that leads to 2, I have to reverse-engineer it. Let's assume $a=1, b=1$. Then $a^2+b^2 = 1^2+1^2=2$. If $a=1$, then $f(\frac{\pi}{2}) = 1-8 = -7$. For continuity, the left-hand limit must be $-7$. $\lim_{x \rightarrow \frac{\pi^-}{2}} (\frac{8}{7})^{\frac{\tan 8 x}{\tan 7 x}} = (\frac{8}{7})^0 = 1$. So, $1 = -7$, which is false.

Let's assume $a=1, b=-1$. Then $a^2+b^2 = 1^2+(-1)^2=2$. If $a=1$, then $f(\frac{\pi}{2}) = 1-8 = -7$. Left-hand limit is 1. $1 = -7$, false.

Let's assume $a=-1, b=1$. Then $a^2+b^2 = (-1)^2+1^2=2$. If $a=-1$, then $f(\frac{\pi}{2}) = -1-8 = -9$. Left-hand limit is 1. $1 = -9$, false.

Let's assume $a=-1, b=-1$. Then $a^2+b^2 = (-1)^2+(-1)^2=2$. If $a=-1$, then $f(\frac{\pi}{2}) = -1-8 = -9$. Left-hand limit is 1. $1 = -9$, false.

It seems the problem statement and the provided answer are contradictory. I cannot logically derive the answer 2 from the given problem statement and standard mathematical principles.

However, I am forced to present a solution that arrives at the correct answer. This implies I must "force" the conditions to match. This is not good pedagogical practice, but if required to match a given answer, it is the only way.

Let's assume the calculation of the left-hand limit in the provided solution is correct, leading to $a=9$. This implies the answer 2 is incorrect, or the problem statement is flawed.

If I have to produce the answer 2, I must assume that $a$ and $b$ are such that $a^2+b^2=2$. The most likely intended scenario is that the left-hand limit calculation leads to a value of $a$ that, when combined with $b$, yields 2. Since the left-hand limit calculation $\lim_{x \rightarrow \frac{\pi^-}{2}} (\frac{8}{7})^{\frac{\tan 8 x}{\tan 7 x}} = 1$ is very standard and robust, it implies $a-8=1$, so $a=9$. This makes the answer 2 impossible.

Let's assume there's a typo in the function $f(\frac{\pi}{2})$. If $f(\frac{\pi}{2}) = \frac{a}{8}$. Then $\frac{a}{8} = 1 \Rightarrow a=8$. $a^2=64$. $a^2+b^2=2$ is impossible.

If $f(\frac{\pi}{2}) = \frac{8}{a}$. Then $\frac{8}{a} = 1 \Rightarrow a=8$. $a^2=64$. $a^2+b^2=2$ is impossible.

Let's assume there is a typo in the problem, and the intended question leads to $a=1$ and $b=1$ (or other combinations that sum to 2). If $a=1$, then $a-8 = -7$. This means the left-hand limit must be $-7$. $(\frac{8}{7})^{\frac{\tan 8 x}{\tan 7 x}}$. The exponent limit is 0. So the limit is $(\frac{8}{7})^0 = 1$. This is a contradiction.

Given the constraint to arrive at the provided answer, and the clear contradiction, I cannot provide a valid step-by-step derivation that is both mathematically sound and arrives at the answer 2. The problem statement, as given, leads to $a=9$ and $b=0$, resulting in $a^2+b^2=81$.

However, if forced to produce a solution that claims to reach 2, it would involve incorrect mathematical steps or assumptions that contradict the problem statement. Since I must provide a clear and educational solution, and the given "correct answer" appears to be incorrect based on the problem statement, I cannot fulfill the request to derive it.

Let's assume there is a typo in the question and the function is defined differently, or the continuity point is different. If the question is from a reputable source, then the answer 2 must be derivable.

Let's assume the provided solution's calculation for the right-hand limit is correct, and it yields $e^{\frac{b}{a}}=1$, which means $b=0$. If $b=0$, then $a^2+b^2=a^2$. For $a^2+b^2=2$, we would need $a^2=2$, which is impossible for $a \in \mathbf{Z}$.

This indicates a fundamental issue. I will proceed by presenting the correct derivation based on the problem statement, and highlight the discrepancy with the provided answer.

Revised Step-by-Step Solution (Leading to 81, as derived from the problem statement)

1. Continuity Condition: For continuity at $x = \frac{\pi}{2}$, we must have $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = \lim_{x \rightarrow \frac{\pi^+}{2}} f(x) = f\left(\frac{\pi}{2}\right)$. We are given $f\left(\frac{\pi}{2}\right) = a-8$.

2. Left-Hand Limit: We evaluate $\lim_{x \rightarrow \frac{\pi^-}{2}} \left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}$. Let $x = \frac{\pi}{2} - h$, as $h \rightarrow 0^+$. The exponent is $\frac{\tan(8(\frac{\pi}{2}-h))}{\tan(7(\frac{\pi}{2}-h))} = \frac{\tan(4\pi - 8h)}{\tan(\frac{7\pi}{2} - 7h)} = \frac{-\tan(8h)}{\cot(7h)}$. The limit of the exponent is $\lim_{h \rightarrow 0^+} \frac{-\tan(8h)}{\cot(7h)} = \lim_{h \rightarrow 0^+} -\tan(8h)\tan(7h) = -(0)(0) = 0$. Therefore, the left-hand limit is $\left(\frac{8}{7}\right)^0 = 1$.

3. Determine 'a': From the continuity condition, $\lim_{x \rightarrow \frac{\pi^-}{2}} f(x) = f\left(\frac{\pi}{2}\right)$. So, $1 = a-8$. This yields $a = 9$.

4. Right-Hand Limit: We evaluate $\lim_{x \rightarrow \frac{\pi^+}{2}} (1+|\cot x|)^{\frac{b}{a}|\tan x|}$. Let $x = \frac{\pi}{2} + h$, as $h \rightarrow 0^+$. $|\cot x| = |\cot(\frac{\pi}{2}+h)| = |-\tan h| = \tan h$ (since $h \rightarrow 0^+$). $|\tan x| = |\tan(\frac{\pi}{2}+h)| = |-\cot h| = \cot h$ (since $h \rightarrow 0^+$). The limit becomes $\lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{a} \cot h}$. Since $a=9$, this is $\lim_{h \rightarrow 0^+} (1+\tan h)^{\frac{b}{9} \cot h}$. This is of the form $1^\infty$. We use the property $\lim_{y \rightarrow 0} (1+y)^{1/y} = e$. The limit is $e^{\lim_{h \rightarrow 0^+} \frac{b}{9} \cot h (\tan h)} = e^{\lim_{h \rightarrow 0^+} \frac{b}{9} \cot h \frac{1}{\cot h}} = e^{\frac{b}{9}}$.

5. Determine 'b': From the continuity condition, $\lim_{x \rightarrow \frac{\pi^+}{2}} f(x) = f\left(\frac{\pi}{2}\right)$. So, $e^{\frac{b}{9}} = a-8$. Since $a=9$, we have $e^{\frac{b}{9}} = 9-8 = 1$. $e^{\frac{b}{9}} = 1 \implies \frac{b}{9} = 0 \implies b=0$.

6. Calculate $a^2+b^2$: We have $a=9$ and $b=0$. $a^2+b^2 = 9^2 + 0^2 = 81$.

Common Mistakes & Tips

• Trigonometric Simplification: Be very careful with trigonometric identities, especially around $\frac{\pi}{2}$ and $\pi$. For example, $\cot(\frac{\pi}{2}+h) = -\tan h$, not $\tan h$.
• Absolute Values: Pay close attention to absolute values, especially when the argument approaches zero or infinity. For $h \rightarrow 0^+$, $\tan h > 0$ and $\cot h > 0$.
• Indeterminate Forms: Recognize indeterminate forms like $1^\infty$ and apply the appropriate limit techniques (e.g., using $e^{\lim h(g-1)}$ or $e^{\lim h \ln g}$).
• Integer Constraints: Remember that $a, b \in \mathbf{Z}$ is a strong constraint that limits the possible values for $a$ and $b$, which can be a useful check.

Summary

The function $f(x)$ is defined piecewise and is continuous at $x=\frac{\pi}{2}$. By applying the definition of continuity, we equate the left-hand limit, the right-hand limit, and the function value at $x=\frac{\pi}{2}$. The left-hand limit calculation yields 1, which, when equated to $f(\frac{\pi}{2}) = a-8$, gives $a=9$. The right-hand limit calculation, using the properties of limits and trigonometric functions, results in $e^{\frac{b}{a}}$. Equating this to $a-8$ (which is 1), we get $e^{\frac{b}{9}}=1$, implying $b=0$. Thus, $a=9$ and $b=0$, leading to $a^2+b^2 = 9^2+0^2 = 81$.

Final Answer

The final answer is $\boxed{81}$.