Find $\lim_{x\to0}(1+x^{2})^{\cot^2{x}}$ without lhopital I need to find $\lim_{x\to 0}(1+x^{2})^{\cot^2{x}}$.

I started like this:

$\lim_{x\to0}(1+x^{2})^{\cot^2{x}} = \lim_{x\to0}e^{\ln(1+x^{2})^{\cot^2{x}}} = \lim_{x\to0}e^{\cot^2{x} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{\frac{\cos^2{x}}{\sin^2{x}} \cdot \ln(1+x^{2})}$

Now I'm not sure about this one but can I say:

$\lim_{x\to0}e^{\frac{\cos^2{x}}{\sin^2{x}} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{\frac{x^2}{\sin^2{x}} \cdot \ln(1+x^{2})}$?

if so than 

$ \lim_{x\to0}e^{\frac{x^2}{\sin^2{x}} \cdot \ln(1+x^{2})} = \lim_{x\to0}e^{1 \cdot \ln(1)} = e^0 = 1$

Is this valid? Is there a better way
No, you can't say that.  $\cos^2 x$ and $x^2$ are not very much alike close to $0$.  The limit is actually $e$.  But your tag says "limits without L'hospital", and I don't see a way to get $e$ without it.
Take the log we get 
$$\cot^2 x \ln(1+x^2)=\cos^2 x \frac{\ln(1+x^2)}{\sin^2 x}$$
write as 

$$\cos^2 x \frac{\ln(1+x^2)}{x^2}\frac{x^2}{\sin^2 x}\to 1$$
by standard limits. Easy peasy.