radicals - Calculating the following limit: $lim_{x to 0} frac{sqrt{x^2+1}-sqrt{x+1}}{1-sqrt{x+1}}$

I am trying to calculate this limit:
$$\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$$

I've tried using conjugate of both denominator and numerator but I can't get the right result.

Answer

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

$$=\frac{(1+\sqrt{x+1})\{x^2+1-(x+1)\}}{(\sqrt{x^2+1}+\sqrt{x+1})(1-(x+1))}$$
$$=\frac{(1+\sqrt{x+1})\{x(x-1)\}}{(\sqrt{x^2+1}+\sqrt{x+1})(-x)}$$
$$=\frac{(1+\sqrt{x+1})(1-x)}{(\sqrt{x^2+1}+\sqrt{x+1})}\text { if } x\ne0$$

As $x\to0,x\ne0$

So, $$\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}=\lim_{x\to0}\frac{(1+\sqrt{x+1})(1-x)}{(\sqrt{x^2+1}+\sqrt{x+1})}=\frac{(1+1)}{(1+1)}=1$$

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Alternatively, as $\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$ is of the form $\frac00,$

Applying L'Hospital's Rule we get,
$$\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$$
$$=\lim_{x\to0}\frac{\frac x{\sqrt{x^2+1}}-\frac1{2\sqrt{x+1}}}{-\frac1{2\sqrt{x+1}}}$$
$$=\lim_{x\to0}\left(1-\frac{2x\sqrt{x+1}}{\sqrt{x^2+1}}\right)\text{ as }x+1\ne0$$
$$=1$$