Suppose $f(x) \rightarrow M$ as $x \rightarrow a$. Prove that if $f(x) \leq L$ for all $x$ near $a$, then $M \leq L$. I'm am a student taking a real analysis paper at university. I'm going through some problems on my problem sheet and I've been asked the question above.
I'm still getting a hang on what it means for $x$ to be "near" $a$ but gather, that if $f(a) \leq L$ and $f(a) = M$ then $M \leq L$ is implied.
If anyone can help me understand the mathematical definition of something being "near" something else and some tips on how to carve a rigorous proof of the above. It would be much appreciated!
Thank you for your time!

Given: $\lim_{x \rightarrow a} f(x)=M$, and in a $\delta$ neighbourhood of $a$, i.e. there is a $\delta >0$ s.t. 
$|x-a| \lt \delta$ implies $f(x)\le L.$
Assume $M >L$.
For $|x-a|<\delta$ we have  $f(x)\le L <M.$
Let $n_0 >1/\delta$ (Archimedean principle).
For $n\ge n_0$
$|x_n-a| <1/n \le 1/n_0 <\delta$, i.e. the $x_n$ are near enough and converge to $a$.
Then
$f(x_n) \le L <M$, and 
$M=\lim_{n \rightarrow \infty}f(x_n)\le L<M$, a contradiction.
