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. $\lim_{x \to c} f(x) = f(c)$. For a piecewise function, this implies that the limit from the left equals the limit from the right, and both are equal to the function's value at that point: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Standard Limits: The following standard limits are useful for evaluating limits of trigonometric functions:
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{\tan x}{x} = 1$
• Taylor Series Expansion: The Taylor series expansion of trigonometric functions around $x=0$ can be used to evaluate indeterminate forms. For small $x$:
  • $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$
  • $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$

Step-by-Step Solution

Step 1: Understand the condition for continuity at x = 0. For the function $f(x)$ to be continuous at $x=0$, the left-hand limit, the right-hand limit, and the function value at $x=0$ must all be equal. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$

Step 2: Calculate the function value at x = 0. The function is defined as $f(x) = a \sin \left( \frac{\pi}{2}(x - 1) \right)$ for $x \le 0$. Substituting $x=0$: $f(0) = a \sin \left( \frac{\pi}{2}(0 - 1) \right) = a \sin \left( -\frac{\pi}{2} \right)$ Since $\sin \left( -\frac{\pi}{2} \right) = -1$, we have: $f(0) = a(-1) = -a$

Step 3: Calculate the left-hand limit at x = 0. The left-hand limit uses the same definition as $f(0)$ since it's for $x \le 0$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} a \sin \left( \frac{\pi}{2}(x - 1) \right)$ As $x \to 0^-$, $(x-1) \to -1$, and $\frac{\pi}{2}(x-1) \to -\frac{\pi}{2}$. $\lim_{x \to 0^-} f(x) = a \sin \left( -\frac{\pi}{2} \right) = a(-1) = -a$ This confirms that $\lim_{x \to 0^-} f(x) = f(0)$.

Step 4: Calculate the right-hand limit at x = 0. The right-hand limit uses the definition $f(x) = \frac{\tan 2x - \sin 2x}{b x^3}$ for $x > 0$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan 2x - \sin 2x}{b x^3}$ This is an indeterminate form of type $\frac{0}{0}$ since $\tan(0)=0$ and $\sin(0)=0$. We can use Taylor series expansions for $\tan(2x)$ and $\sin(2x)$ around $x=0$. Recall the Taylor series: $\sin u = u - \frac{u^3}{3!} + O(u^5)$ $\tan u = u + \frac{u^3}{3} + O(u^5)$ Let $u = 2x$. As $x \to 0$, $u \to 0$. $\sin 2x = 2x - \frac{(2x)^3}{3!} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$ $\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$

Substitute these into the limit expression: $\lim_{x \to 0^+} \frac{\left(2x + \frac{8x^3}{3} + O(x^5)\right) - \left(2x - \frac{4x^3}{3} + O(x^5)\right)}{b x^3}$ $= \lim_{x \to 0^+} \frac{2x + \frac{8x^3}{3} - 2x + \frac{4x^3}{3} + O(x^5)}{b x^3}$ $= \lim_{x \to 0^+} \frac{\frac{12x^3}{3} + O(x^5)}{b x^3}$ $= \lim_{x \to 0^+} \frac{4x^3 + O(x^5)}{b x^3}$ Divide the numerator and denominator by $x^3$: $= \lim_{x \to 0^+} \frac{4 + O(x^2)}{b}$ As $x \to 0$, $O(x^2) \to 0$. $\lim_{x \to 0^+} f(x) = \frac{4}{b}$

Step 5: Equate the limits and function value for continuity. From Step 2 and Step 4, for continuity at $x=0$: $\lim_{x \to 0^+} f(x) = f(0)$ $\frac{4}{b} = -a$ This gives us the relationship between $a$ and $b$: $ab = -4$

Step 6: Calculate the required expression. We need to find the value of $10 - ab$. Substitute the value of $ab$ we found: $10 - ab = 10 - (-4)$ $10 - ab = 10 + 4$ $10 - ab = 14$

Common Mistakes & Tips

• Incorrect Taylor Series Expansion: Ensure you use the correct terms in the Taylor series expansion. For $\tan u$, the first two terms are $u + \frac{u^3}{3}$, and for $\sin u$, they are $u - \frac{u^3}{6}$.
• Forgetting the Function Value: The continuity condition requires all three parts (left limit, right limit, and function value) to be equal. While the left limit and function value were equal in this case, it's a good habit to calculate all three explicitly.
• Algebraic Errors in Limit Evaluation: Be meticulous with algebraic manipulations, especially when simplifying the expression after substituting the Taylor series. Dividing by $x^3$ correctly is crucial.

Summary

To ensure continuity of the function $f(x)$ at $x=0$, we equated the left-hand limit, the right-hand limit, and the function's value at $x=0$. We calculated $f(0)$ and the left-hand limit using the first part of the piecewise definition, which yielded $-a$. For the right-hand limit, we used the second part of the definition and employed Taylor series expansions for $\tan(2x)$ and $\sin(2x)$ to resolve the indeterminate form $\frac{0}{0}$. This evaluation resulted in $\frac{4}{b}$. Equating $\frac{4}{b}$ to $-a$ gave us the product $ab = -4$. Finally, we substituted this value into the expression $10 - ab$ to find the answer.

The final answer is \boxed{14}.