I have been trying to understand this limit:

$$\lim_{x \to 0}\frac{tan(x)-sin(x)}{x^2}$$

When aplying the l'Hopital rule I arrive to the limit being $0$ but when doing things organically I get an indetermination:

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

Clearly $\lim_{x \to 0} \frac{1}{cos(x)}=1$ hence $(\frac{1}{cos(x)}-1)=0$ and I could well aply $\lim_{x \to 0}\frac{sin(x)}{x}=1$ but that still leaves $\lim_{x \to 0}\frac{1}{x}$ which is undetermined because it has different limits on $0^-$ and $0^+$.

Is there something I'm missing?


1 Answer 1


Rearrange this as $\frac{\sin x}{x} \frac{1}{\cos x}\frac{1- \cos x}{x}$ and use the standard limit $\frac{1- \cos x}{x} \to 0$ as $x \to 0$.

  • $\begingroup$ Thankyou that is exactly what I was missing! $\endgroup$
    – Rocketman
    Commented Nov 1, 2020 at 1:55
  • $\begingroup$ @Rocketman: You’re welcome. $\endgroup$
    – RRL
    Commented Nov 1, 2020 at 2:51

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