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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:

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

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

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

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

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

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

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

tryingtobeastoic
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    Error In the last step, the limit is in 'indeterminant form' that is $\infty. 0 $ and does not equal to $0.$ – Manish Saini Oct 02 '21 at 06:46
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    @user10354138 Iìm not sure that the target duplicate is an exact duplicate for this question. Here the OP is asking for the mistake in their derivation. Moreover the OP is also already showing a way to solve a similar problem which is almost a solution for the presented one. – user Oct 02 '21 at 06:53
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    @user10354138 I agree that the one you selected is strictly related but it doesn't address the question posed by the OP. For this reason I'm going to vote to reopen it. – user Oct 02 '21 at 06:55

3 Answers3

3

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

and to the related

tryingtobeastoic
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user
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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}

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

$ \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)$

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

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

Lai
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