Can the limits and derivatives of a function always be solved by applying its basic theorems? I'm studying calculus, specifically about the limits and derivatives of a function. So I became interested in studying the basic theorems and the properties of limits and derivatives of a function. However, throughout the theory, the following question arises:
Can the limits and derivatives of a function always be solved by applying its basic theorems?
I would like to know if this is true or false and I think the best way to do it is through an example, and I am looking for it.
 A: I suspect the OP has something simpler than the comments in mind. Consider the function
$$
f(x)
=
\begin{cases}
x\sin \left(\frac{1}{x}\right) & x \neq 0 \\
0 & x = 0
\end{cases}
$$
You can show that $\lim_{x \to 0} f(x) = 0$, but $f'(0)$ does not exist.
By contrast
$$
g(x)
=
\begin{cases}
x^2\sin \left(\frac{1}{x}\right) & x \neq 0 \\
0 & x = 0
\end{cases}
$$
has $g'(0) = 0$.
You can't use the "basic theorems" to compute these limits and derivatives. You have to make some kind of special argument. For these functions, the squeeze theorem is probably the most expedient route, but you could also argue from the definition.
There are other more difficult functions to compute the limits and derivatives of, but hopefully this gives you some idea that computing limits and derivatives is not always a trivial exercise. Indeed, there are cases where the best we can do is say that the limit exists and then approximate it with numerical methods.
A: No. There are another theorems which are necessary for indeterminate limits.
For example the classical limit
$$ \lim_{h\to 0} \frac{\sin(h)}{h}.$$
It can't be solved using only the basic theorems. In your course you will learn a specific proof that demostrates that this limit equals 1.
