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:

$\underset{x\to 0}{lim}\frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}}$

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

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

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

0

However, according to wolframalpha and my book, I'm wrong.

My attempt:

0

However, according to wolframalpha and my book, I'm wrong.

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

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

We have that

veiga34

Beginner2021-12-31Added 32 answers

The error lies in the equality

You cannot take the limit at 0 in part of your expression and leave the x in the remaining expression.

You have

nick1337

Expert2022-01-08Added 699 answers

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