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?