How do you find the limit of $\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}$?

Question

My attempt:

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

$$\lim_{x\to 0}\frac{1}{x^2}(\frac{\tan x}{x}-\frac{\sin x}{x})$$

$$\lim_{x\to 0}\frac{1}{x^2}(1-1)$$

$$\lim_{x\to 0}\frac{1}{x^2}\times 0$$

$$0$$

However, according to wolframalpha and my book, I'm wrong. The correct answer is $\frac{1}{2}$.

Questions:

1. Why did I get the wrong answer in my process?
2. What would be the correct way to solve the problem?

Solution to a similar problem:

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

$$\lim_{x\to 0}\frac{\frac{\sin x}{\cos x}-\sin x}{\sin^3x}$$

$$\lim_{x\to 0}\frac{\frac{\sin x-\cos x.\sin x}{\cos x}}{\sin^3x}$$

$$\lim_{x\to 0}{\frac{\sin x-\cos x.\sin x}{\cos x.\sin^3x}}$$

$$\lim_{x\to 0}\frac{(1-\cos x)}{\cos x.\sin^2x}$$

$$\lim_{x\to 0}\frac{(1-\cos x)}{\cos x(1+\cos x)(1-\cos x)}$$

$$\frac{1}{2}$$

Answer 1

Your way is wrong because you are taking the limit not at once for the whole expression and this is not allowed in general.

We have that

$$\frac{\tan x-\sin x}{x^3}=\frac{\tan x-\sin x}{\sin^3x}\frac{\sin^3x}{x^3}$$

then we can use the second solution you have shown.

For the mistake in the first solution refer also to

• Analyzing limits problem Calculus (tell me where I'm wrong).

and to the related

• Trig limit without L'Hospital Rule: $\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}$

Answer 2

The error lies in the equality$$\lim_{x\to0}\frac1{x^2}\left(\frac{\tan x}x-\frac{\sin x}x\right)=\lim_{x\to0}\frac1{x^2}(1-1).$$You cannot take the limit at $0$ in part of your expression and leave the $x$ in the remaining expression.

You have\begin{align}\require{cancel}\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}&=\lim_{x\to0}\frac{\left(\cancel x+\frac13x^3+O(x^4)\right)-\left(\cancel x-\frac16x^3+O(x^4)\right)}{x^3}\\&=\lim_{x\to0}\frac{\frac12x^3+O(x^4)}{x^3}\\&=\frac12.\end{align}

Answer 3

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}}\left(\frac{0}{0}\right) \\=& \lim _{x \rightarrow 0} \frac{\sec ^{2} x-\cos x}{3 x^{2}}\left(\frac{0}{0}\right) \\=& \lim _{x \rightarrow 0} \frac{2 \sec ^{2} x \tan x+\sin x}{6 x} \end{aligned}$

$ \displaystyle =\frac{1}{6}\left(2 \lim _{x \rightarrow 0} \sec ^{2} x \cdot \lim _{x \rightarrow 0} \frac{\tan x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)$

$\displaystyle =\frac{1}{6}(2 \cdot 1+1)$

$\displaystyle =\frac{1}{2}$