Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Limit of an Exponential Function: If $\lim_{x \to c} g(x) = L$ and $\lim_{x \to c} h(x) = M$, then $\lim_{x \to c} [g(x)]^{h(x)} = L^M$, provided $L^M$ is well-defined.
• Standard Limit: $\lim_{x \to 0} \frac{\tan(ax)}{bx} = \frac{a}{b}$ and $\lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} = \frac{a}{b}$.

Step-by-Step Solution

Step 1: Understand the Condition for Continuity The problem states that the function $f(x)$ is continuous at $x = \frac{\pi}{2}$. For continuity at $x = \frac{\pi}{2}$, we must have $\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$.

Step 2: Determine the Value of $f\left(\frac{\pi}{2}\right)$ From the definition of the function, when $x = \frac{\pi}{2}$, $f(x) = k + \frac{2}{5}$. So, $f\left(\frac{\pi}{2}\right) = k + \frac{2}{5}$.

Step 3: Calculate the Limit of $f(x)$ as $x \to \frac{\pi}{2}$ We need to find $\lim_{x \to \frac{\pi}{2}} f(x)$, where $f(x) = \left(\frac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}}$ for $0 < x < \frac{\pi}{2}$. $\lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \left(\frac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}}$ This is an exponential function. We can evaluate the limit of the exponent first. Let $g(x) = \frac{\tan 4x}{\tan 5x}$.

Step 4: Evaluate the Limit of the Exponent We need to evaluate $\lim_{x \to \frac{\pi}{2}} \frac{\tan 4x}{\tan 5x}$. As $x \to \frac{\pi}{2}$, $4x \to 2\pi$ and $5x \to \frac{5\pi}{2}$. The tangent function has a period of $\pi$. So, $\tan(4x) = \tan(4x - 2\pi)$ and $\tan(5x) = \tan(5x - 2\pi) = \tan(5x - \frac{4\pi}{2})$. Let $y = x - \frac{\pi}{2}$. As $x \to \frac{\pi}{2}$, $y \to 0$. Then $x = y + \frac{\pi}{2}$. $4x = 4\left(y + \frac{\pi}{2}\right) = 4y + 2\pi$ $5x = 5\left(y + \frac{\pi}{2}\right) = 5y + \frac{5\pi}{2}$ Now, substitute these into the exponent: $\frac{\tan 4x}{\tan 5x} = \frac{\tan(4y + 2\pi)}{\tan(5y + \frac{5\pi}{2})}$ Using the periodicity of tangent, $\tan(\theta + n\pi) = \tan(\theta)$ for integer $n$, and $\tan(\theta + \frac{\pi}{2}) = -\cot(\theta)$. $\tan(4y + 2\pi) = \tan(4y)$ $\tan(5y + \frac{5\pi}{2}) = \tan(5y + 2\pi + \frac{\pi}{2}) = \tan(5y + \frac{\pi}{2}) = -\cot(5y)$ So, the exponent becomes: $\frac{\tan(4y)}{-\cot(5y)} = -\frac{\tan(4y)}{\cot(5y)} = -\tan(4y) \tan(5y)$ Now, we evaluate the limit as $y \to 0$: $\lim_{y \to 0} (-\tan(4y) \tan(5y))$ We can rewrite this as: $- \lim_{y \to 0} \left(\frac{\tan(4y)}{y} \cdot \frac{\tan(5y)}{y} \cdot y^2\right)$ Using the standard limit $\lim_{z \to 0} \frac{\tan(az)}{z} = a$: $- \left(\lim_{y \to 0} \frac{\tan(4y)}{y}\right) \cdot \left(\lim_{y \to 0} \frac{\tan(5y)}{y}\right) \cdot \left(\lim_{y \to 0} y^2\right)$ $- (4) \cdot (5) \cdot (0) = 0$ Alternatively, we can use the limit $\lim_{x \to c} \frac{\tan(ax)}{\tan(bx)} = \frac{a}{b}$ by making a substitution. Let $z = x - \frac{\pi}{2}$. As $x \to \frac{\pi}{2}$, $z \to 0$. Then $4x = 4(z + \frac{\pi}{2}) = 4z + 2\pi$, and $5x = 5(z + \frac{\pi}{2}) = 5z + \frac{5\pi}{2}$. $\lim_{x \to \frac{\pi}{2}} \frac{\tan 4x}{\tan 5x} = \lim_{z \to 0} \frac{\tan(4z + 2\pi)}{\tan(5z + \frac{5\pi}{2})} = \lim_{z \to 0} \frac{\tan(4z)}{\tan(5z + \frac{\pi}{2})}$ $= \lim_{z \to 0} \frac{\tan(4z)}{-\cot(5z)} = \lim_{z \to 0} \frac{\sin(4z)}{\cos(4z)} \cdot (-\sin(5z))$ This approach seems to lead to a $0/0$ indeterminate form. Let's go back to the substitution $y = x - \frac{\pi}{2}$. We had the exponent as $-\tan(4y)\tan(5y)$. $\lim_{y \to 0} (-\tan(4y) \tan(5y))$ This is of the form $0 \times 0$, which is 0. The limit of the exponent is 0.

Step 5: Calculate the Limit of $f(x)$ using the Limit of the Exponent Now we can find the limit of the function: $\lim_{x \to \frac{\pi}{2}} f(x) = \left(\frac{4}{5}\right)^{\lim_{x \to \frac{\pi}{2}} \frac{\tan 4x}{\tan 5x}} = \left(\frac{4}{5}\right)^0$ Any non-zero number raised to the power of 0 is 1. $\lim_{x \to \frac{\pi}{2}} f(x) = 1$

Step 6: Equate the Limit and the Function Value to Find $k$ For continuity, we have $\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$. $1 = k + \frac{2}{5}$ Now, solve for $k$: $k = 1 - \frac{2}{5}$ $k = \frac{5}{5} - \frac{2}{5}$ $k = \frac{3}{5}$

Common Mistakes & Tips

• Incorrectly evaluating the limit of the exponent: A common mistake is to directly substitute $x = \frac{\pi}{2}$ into $\tan(4x)$ and $\tan(5x)$, which leads to $\tan(2\pi)$ and $\tan(\frac{5\pi}{2})$, both of which are undefined in their direct form. Proper substitution or use of limit properties is crucial.
• Misapplication of standard limits: The standard limit $\lim_{x \to 0} \frac{\tan(ax)}{bx} = \frac{a}{b}$ is useful here after a suitable substitution to transform the limit to $x \to 0$.
• Algebraic errors: Ensure careful algebraic manipulation when solving for $k$ after equating the limit and function value.

Summary

To ensure the function $f(x)$ is continuous at $x = \frac{\pi}{2}$, we equated the limit of the function as $x$ approaches $\frac{\pi}{2}$ with the value of the function at $x = \frac{\pi}{2}$. We calculated the limit of the function by first finding the limit of its exponent, $\frac{\tan 4x}{\tan 5x}$, as $x \to \frac{\pi}{2}$. Through a substitution $y = x - \frac{\pi}{2}$ and using the periodicity of the tangent function, we found the limit of the exponent to be 0. This resulted in the limit of $f(x)$ being $\left(\frac{4}{5}\right)^0 = 1$. By setting this equal to $f\left(\frac{\pi}{2}\right) = k + \frac{2}{5}$, we solved for $k$ and found it to be $\frac{3}{5}$.

The final answer is $\boxed{\frac{3}{5}}$.