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: $\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}$