Question

Evaluate the following limit if it exists:
limx→0 [4x^3]/[tan^(3) x+tanx-x]

Answer 1

Answer on Question #69104 – Math – Real Analysis

Question

Evaluate the following limit if it exists:

$$\lim_{x \to 0} \frac{4x^3}{\tan^3 x + \tan x - x}$$

Solution

$$\begin{aligned} \lim_{x \to 0} \frac{4x^3}{\tan^3 x + \tan x - x} &= \lim_{x \to 0} \frac{4x^3}{\tan x (\tan^2 x + 1) - x} = \lim_{x \to 0} \frac{4x^3}{\frac{\tan x}{\cos^2 x} - x} = [L' \text{Hopital's rule}] = \\ &= \lim_{x \to 0} \frac{(4x^3)'}{\left(\frac{\tan x}{\cos^2 x} - x\right)'} = \lim_{x \to 0} \frac{12x^2}{\frac{1}{\cos^4 x} + \frac{2 \tan x \cdot \sin x}{\cos^3 x} - 1} = [L' \text{Hopital's rule}] = \\ &= \lim_{x \to 0} \frac{(12x^2)'}{\left(\frac{1}{\cos^4 x} + \frac{2 \tan^2 x}{\cos^2 x} - 1\right)'} = \lim_{x \to 0} \frac{24x}{\frac{4 \sin x}{\cos^5 x} + \frac{4 \tan x}{\cos^4 x} + \frac{4 \tan^2 x \cdot \sin x}{\cos^3 x}} = [L' \text{Hopital's rule}] \\ &= \lim_{x \to 0} \frac{(24x)'}{\left(\frac{8 \tan x}{\cos^4 x} + \frac{4 \tan^3 x}{\cos^2 x}\right)'} = \lim_{x \to 0} \frac{24}{\frac{8}{\cos^6 x} + \frac{32 \tan x \cdot \sin x}{\cos^5 x} + \frac{12 \tan^2 x}{\cos^4 x} + \frac{4 \tan^3 x \cdot \sin x}{\cos^3 x}} \\ &= \frac{24}{8 + 0 + 0 + 0} = 3 \end{aligned}$$

Answer: 3.

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