Stuck with homework (limit exercise) - squeeze theorem Given function $f:\mathbb R→\mathbb R$ for which $|f(x)-2|≤x^2$.
Find the limits:

$\lim_{x\to0}f(x)$.
$\lim_{x\to0}\frac{f(x)-\sqrt{x+4}}{x}$.


I can solve question (a) very easily (squeeze theorem) and the answer is $\lim_{x \rightarrow 0} f(x)= 2$.
But when I start solving question (b), I'm in a dead end. The book says that the answer of question (b) is $\lim_{x \rightarrow 0 }\frac {f(x)-\sqrt{x+4}}{x} = -\frac{1}{4}$.
Perhaps I'm doing an algebra mistake somewhere because I end up always at $-\frac{1}{0}$ which is not possible. Could someone please give me a hint ?
Thanks
You can split your limit into two parts:

Part 1: $\frac{f(x)-2}{x}$ which tends to zero.

Part 2: $\frac{2-\sqrt{x+4}}{x}$ which tends to $-1/4$ by L'Hopital's rule.
a) $lim f(x)$ for $x→0 =2$
because 

$0&lt;=|f(x)-2|&lt;=x^2$, by squeeze theorem, 

as $x$ goes to $0,0&lt;=|f(x)-2|&lt;=0.
so |f(x)-2|=0$, 

hence $lim f(x)$ for $x→0 =2$