Evaluate the limit: lim (tan(2x) - sin(2x)) / x3 as x approaches 0

Question

ScanSolve AI Tutor · Accepted Answer

To evaluate the limit, we first substitute $x=0$ into the expression:

\frac{\tan(2 \cdot 0) - \sin(2 \cdot 0)}{0^3} = \frac{\tan(0) - \sin(0)}{0} = \frac{0 - 0}{0} = \frac{0}{0}

This is an indeterminate form, so we need to manipulate the expression.

Rewrite $\tan(2x)$ in terms of $\sin(2x)$ and $\cos(2x)$ .
$\lim_{x \to 0} \frac{\frac{\sin(2x)}{\cos(2x)} - \sin(2x)}{x^3}$
Factor out $\sin(2x)$ from the numerator and combine the terms.
$\lim_{x \to 0} \frac{\sin(2x) \left( \frac{1}{\cos(2x)} - 1 \right)}{x^3} = \lim_{x \to 0} \frac{\sin(2x) \left( \frac{1 - \cos(2x)}{\cos(2x)} \right)}{x^3}$
Rearrange the terms to utilize known standard limits.
$\lim_{x \to 0} \left( \frac{\sin(2x)}{x} \cdot \frac{1 - \cos(2x)}{x^2} \cdot \frac{1}{\cos(2x)} \right)$
Apply the standard limits
- $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$
- $\lim_{x \to 0} \frac{1 - \cos(ax)}{x^2} = \frac{a^2}{2}$
- $\lim_{x \to 0} \frac{1}{\cos(ax)} = \frac{1}{\cos(0)} = 1$
Applying these to our expression:
$\left( \lim_{x \to 0} \frac{\sin(2x)}{x} \right) \cdot \left( \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \right) \cdot \left( \lim_{x \to 0} \frac{1}{\cos(2x)} \right)$ $= (2) \cdot \left( \frac{2^2}{2} \right) \cdot (1)$ $= 2 \cdot \left( \frac{4}{2} \right) \cdot 1$ $= 2 \cdot 2 \cdot 1$ $= 4$
The limit exists and its value is:
$4$

Answer