Existence of a limit Prove if the following is correct or not: If $$\lim _{x\to x_0}f(x) = L \text{ and } \lim_{x\to x_1}g(x) = x_0,$$ then $$\lim_{x\to x_1} f(g(x))= L.$$
So, I guess this can be solved either by proving it or find a example that contradicts the above, so this statement is wrong.
 A: For all $\varepsilon>0$ there exists $\eta>0$ such that whenever $0<|x-x_0|<\eta$ then $|f(x)-L|<\varepsilon$.
Given $\eta>0$, there exists $\delta>0$ such that whenever $0<|x-x_1|<\delta$ then $|g(x) - x_0|<\eta$.
So $\underbrace{0<|x-x_1| <\delta \Longrightarrow |g(x)-x_0|<\eta \Longrightarrow |f(g(x)) - L|<\varepsilon}_\text{This  falls just a little bit short.}$.
The difficulty here is that we don't have $0<|g(x) -x_0|<\eta$.  That suggests that if $g(x) = x_0$, there might be a problem.  Consider the case where $g(x) = x_0$ regardless of the value of $x$.  Then $g$ is a constant function.  Suppose then that $f(x) = \dfrac{x^2 - x_0^2}{x-x_0}$.  This simplifies to $x+x_0$ when $x\ne x_0$, but is undefined when $x=x_0$. It limit as $x\to x_0$ is $L=2x_0$.  If $g$ is constantly equal to $x_0$, then $\lim_{x\to x_1} g(x) = x_0$ but $f(g(x))$ would be undefined, and thus cannot approach $L$.
However, if $f$ is continuous at $x_0$, then the conclusion is correct and the argument above is almost valid.  The only thing one would need to change to make it valid is to say that if $f$ is continuous at $x_0$ and $\lim_{x\to x_0} f(x) = L$ (which, in this case of continuity, implies $f(x_0)=L$), is that we would need to say the following:
$$
\text{For all }\varepsilon>0 \text{ there exists } \eta>0 \text{ such that whenever } |x-x_0|<\eta \text{ then } |f(x)-L|<\varepsilon.
$$
