Prove that $\lim_{x \rightarrow 0} \frac{\tan(x)-\sin(x)}{\sin^3(x)} = \frac{1}{2}$

Let’s first expand the $\tan(x)$ into $\frac{\sin(x)}{\cos(x)}$.

$$ \lim_{x \rightarrow 0} \frac{\frac{\sin(x)}{\cos(x)}-\sin(x)}{\sin^3(x)} $$

Now, we’ll simplify the fraction by simplifying the numerator.

$$ \lim_{x \rightarrow 0} \frac{\sin(x) - \sin(x)\cos(x)}{\cos(x)\sin^3(x)} $$

We can cancel out one of the $\sin(x)$s by taking it common in the numerator.

$$ \lim_{x \rightarrow 0} \frac{\sin(x)(1 - \cos(x))}{\cos(x)\sin^3(x)} $$

So now we have:

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

The real trickery is finding out what to do now. So far, I don’t think there’s any way to just “know” what to do at this stage, so you’re gonna have to do some trial and error. I found, after some time, that multiplying by the conjugate of the numerator helped!

$$ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{\cos(x)\sin^2(x)} \times \frac{1+\cos(x)}{1+\cos(x)} $$

We’ll expand out the whole thing now:

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

Using the much loved trigonometric identity of $\sin^2(x) + \cos^2(x) = 1$ and hence $\sin^2(x) = 1 - \cos^2(x)$, we end up with $\sin^2(x)$ in the numerator.

$$ \lim_{x \rightarrow 0} \frac{\sin^2(x)}{\cos(x)\sin^2(x) \cos^2(x)\sin^2(x)} $$

We’ll now cancel out all of the $\sin^2(x)$s, so we end up with:

$$ \lim_{x \rightarrow 0} \frac{1}{\cos(x) + \cos^2(x)} $$

If we “apply” the limit,

$$ \frac{1}{\cos(0) + \cos^2(0)} = \frac{1}{1+1} = \frac{1}{2}. $$

Cool!