Understanding Limit without l'Hopital

Question

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?

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 $x \to 0$.

Answer 2

In the numerator, since both $\tan(x)$ and $\sin(x)$ are odd functions, the difference is also an odd function. The denominator $x^2$ is an even function. The quotient is an odd function. If the limit as $x$ approaches $0$ from positive reals exists, then it is the negative of the limit from negative reals. Thus, the bidirectional limit exists and it must be equal to its negative. That implies the limit is zero if it exists.

Your last step is not valid since the first factor goes to infinity and the second goes to zero. This is an indeterminate form and hence we can't tell what the limit is from that.