Solving a limit with radicals without l'Hopital

I've been trying to solve this particular expression, rationalizing the numerator, and the denominator by conjugate multiplying, squaring, multiplying/dividing with x/x, nothing seems to work, I would appreciate any input.

$$\lim_{x\to 0} \frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1+x}-1}$$

Answer

\begin{align}
\frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1+x}-1}
&=
\frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1+x}-1}
\frac{\sqrt{1+x}+\sqrt{1+x^2}}{\sqrt{1+x}+1}
\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1+x^2}}\\[6px]
&=
\frac{(1+x)-(1+x^2)}{(1+x)-1}

\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1+x^2}}\\[6px]
&=
\frac{x(1-x)}{x}
\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1+x^2}}\\[6px]
\end{align}
Now it's easy, isn't it?