Susan Nall

## Answered question

2021-12-29

How do you find the limit of $\underset{x\to 0}{lim}\frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}}$?
My attempt:
$\underset{x\to 0}{lim}\frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}}$
$\underset{x\to 0}{lim}\frac{1}{{x}^{2}}\left(\frac{\mathrm{tan}x}{x}-\frac{\mathrm{sin}x}{x}\right)$
$\underset{x\to 0}{lim}\frac{1}{{x}^{2}}\left(1-1\right)$
$\underset{x\to 0}{lim}\frac{1}{{x}^{2}}×0$
0
However, according to wolframalpha and my book, I'm wrong.

### Answer & Explanation

twineg4

Beginner2021-12-30Added 33 answers

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

veiga34

Beginner2021-12-31Added 32 answers

The error lies in the equality
$\underset{x\to 0}{lim}\frac{1}{{x}^{2}}\left(\frac{\mathrm{tan}x}{x}-\frac{\mathrm{sin}x}{x}\right)=\underset{x\to 0}{lim}\frac{1}{{x}^{2}}\left(1-1\right)$
You cannot take the limit at 0 in part of your expression and leave the x in the remaining expression.
You have
$\underset{x\to 0}{lim}\frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}}=\underset{x\to 0}{lim}\frac{\left(x+{\frac{13}{x}}^{3}+O\left({x}^{4}\right)\right)-\left(x-{\frac{16}{x}}^{3}+O\left({x}^{4}\right)\right)}{{x}^{3}}$
$=\underset{x\to 0}{lim}\frac{{\frac{12}{x}}^{3}+O\left({x}^{4}\right)}{{x}^{3}}$
$=\frac{12}{}$

nick1337

Expert2022-01-08Added 777 answers

$\begin{array}{}\underset{x\to 0}{lim}\frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}}\left(\frac{0}{0}\right)\\ =\underset{x\to 0}{lim}\frac{{\mathrm{sec}}^{2}x-\mathrm{cos}x}{3{x}^{2}}\left(\frac{0}{0}\right)\\ =\underset{x\to 0}{lim}\frac{2{\mathrm{sec}}^{2}x\mathrm{tan}x+\mathrm{sin}x}{6x}\\ =\frac{1}{6}\left(2\cdot 1+1\right)\\ =\frac{1}{2}\end{array}$

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