In Apostol calculus exercise there is a bunch of exercises where we need to find a limit of a composite function. One example is this:
$\lim_\limits{x \rightarrow \frac{\pi}{2}} \frac{\sin(\cos(x))}{\cos(x)}$
Well, intuitively we simply take $\cos(x) \rightarrow 0$ as $x \rightarrow \pi/2$. Then $\lim_\limits{x \rightarrow \frac{\pi}{2}} \frac{\sin(\cos(x))}{\cos(x)} = 1$, since the limit of $\sin(x)/x$ is 1 as $x \rightarrow 0$. Multiple online solutions suggest the same without any explanation.
How to show this rigorously? In Apostol, there is a proof that composition of two continuous functions is continuous at any $x = p$. However, there is no proof about general limits. Moreover, we know that $\sin(x)/x$ is not even defined at $0$. How is it allowed to substitute $\cos(x) \rightarrow 0$ into this outer sine then?
PS. L'Hopital rule has not been specified yet. The only available option is delta-epsilon arguments, and simplest limit rules: sum, difference, product, and division. Squeezing limit theorem can be used (proved before). Plus sin(x)/x limit at 0 has been proven with squeezing limit theorem.
OK, then you're basically done. Fix $\epsilon>0$. Choose $\delta$ so that $|(\sin y)/y-1 | < \epsilon$ if $0<|y|<\delta$. Now choose $0<\delta_1<1$ such that $$ 0<|x-\pi/2| < \delta_1 \implies |\cos x | < \delta.$$ This implies $$ 0<|x-\pi/2| < \delta_1 \implies \left|\frac{\sin\cos x }{ \cos x} - 1\right| < \epsilon.$$
Further,
You will find that the proof for limits is nearly the same as the proof for continuous functions, I recommend it as an exercise for you.