Evaluate the limit: lim (x-0) [tan(3x) − 3tan(x)] / x3

Question

Evaluate the limit: lim (x->0) [tan(3x) − 3tan(x)] / x3

ScanSolve AI Tutor · Accepted Answer

Okay, ~Mataa🫶🏽, let's tackle the second limit.

First, substitute $x=0$ into the expression to check for an indeterminate form:

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

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

Use the triple angle identity for tangent, $\tan(3x) = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}$ .
Substitute this into the limit expression:
$\lim_{x \to 0} \frac{\frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} - 3 \tan x}{x^3}$
Combine the terms in the numerator by finding a common denominator.
$\lim_{x \to 0} \frac{3 \tan x - \tan^3 x - 3 \tan x (1 - 3 \tan^2 x)}{x^3 (1 - 3 \tan^2 x)}$ $\lim_{x \to 0} \frac{3 \tan x - \tan^3 x - 3 \tan x + 9 \tan^3 x}{x^3 (1 - 3 \tan^2 x)}$
Simplify the numerator.
$\lim_{x \to 0} \frac{8 \tan^3 x}{x^3 (1 - 3 \tan^2 x)}$
Rearrange the terms to utilize the standard limit $\lim_{x \to 0} \frac{\tan x}{x} = 1$ .
$\lim_{x \to 0} \left( 8 \cdot \left( \frac{\tan x}{x} \right)^3 \cdot \frac{1}{1 - 3 \tan^2 x} \right)$
Evaluate each part of the limit.
$\left( \lim_{x \to 0} 8 \right) \cdot \left( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^3 \right) \cdot \left( \lim_{x \to 0} \frac{1}{1 - 3 \tan^2 x} \right)$ $= 8 \cdot (1)^3 \cdot \frac{1}{1 - 3 \tan^2(0)}$ $= 8 \cdot 1 \cdot \frac{1}{1 - 3 \cdot 0^2}$ $= 8 \cdot 1 \cdot \frac{1}{1 - 0}$ $= 8 \cdot 1 \cdot 1$ $= 8$
The limit exists and its value is:
$8$

Evaluate the limit: lim (x->0) [tan(3x) - 3tan(x)] / x3