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

Question

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 1

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

As $x\to0,x\ne0$

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

---

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, $$\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$$ $$=\lim_{x\to0}\frac{\frac x{\sqrt{x^2+1}}-\frac1{2\sqrt{x+1}}}{-\frac1{2\sqrt{x+1}}}$$ $$=\lim_{x\to0}\left(1-\frac{2x\sqrt{x+1}}{\sqrt{x^2+1}}\right)\text{ as }x+1\ne0$$ $$=1$$

Answer 2

$$ \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}=\frac{x^2-x}{-x}\cdot\frac{1+\sqrt{x+1}}{\sqrt{x^2+1}+\sqrt{x+1}}\sim1\cdot\frac{2}{2} $$