Susan Nall

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.

twineg4

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

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