Question

$\lim_{x \to 0} \frac{\sin 2x + a \sin x}{x^3} = L$. If $L$ is finite, then $L - a =$

Answer 1

1

Answer 2

The problem asks us to find the value of $L-a$ given that $\lim_{x \to 0} \frac{\sin 2x + a \sin x}{x^3} = L$ and $L$ is finite.

1. Analyze the Indeterminate Form: As $x \to 0$, the numerator $\sin 2x + a \sin x \to \sin(0) + a \sin(0) = 0$. The denominator $x^3 \to 0$. So, the limit is of the indeterminate form $\frac{0}{0}$.

2. Use Taylor Series Expansion: To evaluate the limit and find the value of $a$ that makes $L$ finite, we can use the Taylor series expansion for $\sin x$ and $\sin 2x$ around $x=0$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + O(x^5)$ $\sin 2x = (2x) - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \dots = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$

3. Substitute Expansions into the Numerator: Substitute these expansions into the numerator of the given limit expression: Numerator ($N(x)$) $= \sin 2x + a \sin x$ $N(x) = \left(2x - \frac{4x^3}{3} + O(x^5)\right) + a \left(x - \frac{x^3}{6} + O(x^5)\right)$ Group terms by powers of $x$: $N(x) = (2+a)x - \left(\frac{4}{3} + \frac{a}{6}\right)x^3 + O(x^5)$

4. Determine the value of 'a' for a Finite Limit: For the limit $L = \lim_{x \to 0} \frac{N(x)}{x^3}$ to be finite, the lowest power of $x$ in the numerator must be at least $x^3$. This means the coefficient of $x$ in $N(x)$ must be zero. $2+a = 0$ $a = -2$

5. Calculate the value of 'L': Substitute $a = -2$ back into the expression for $N(x)$: $N(x) = (2 + (-2))x - \left(\frac{4}{3} + \frac{-2}{6}\right)x^3 + O(x^5)$ $N(x) = 0x - \left(\frac{4}{3} - \frac{1}{3}\right)x^3 + O(x^5)$ $N(x) = -\left(\frac{3}{3}\right)x^3 + O(x^5)$ $N(x) = -x^3 + O(x^5)$

Now, substitute this back into the limit expression for $L$: $L = \lim_{x \to 0} \frac{-x^3 + O(x^5)}{x^3}$ $L = \lim_{x \to 0} \left(\frac{-x^3}{x^3} + \frac{O(x^5)}{x^3}\right)$ $L = \lim_{x \to 0} (-1 + O(x^2))$ As $x \to 0$, $O(x^2) \to 0$. So, $L = -1$.

6. Calculate L - a: We found $a = -2$ and $L = -1$. $L - a = -1 - (-2)$ $L - a = -1 + 2$ $L - a = 1$

The final answer is $\boxed{1}$.

Explanation of the solution: The limit is of the form $\frac{0}{0}$. For the limit to be finite, the lowest power of $x$ in the numerator must be at least $x^3$. Using Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \dots$ and $\sin 2x = 2x - \frac{4x^3}{3} + \dots$, the numerator becomes $(2+a)x - (\frac{4}{3} + \frac{a}{6})x^3 + \dots$. For the limit to be finite, the coefficient of $x$ must be zero, so $2+a=0 \implies a=-2$. Substituting $a=-2$, the numerator becomes $-x^3 + \dots$. Dividing by $x^3$, the limit $L = -1$. Therefore, $L-a = -1 - (-2) = 1$.