How do you find #lim (sqrt(x^2+1)-1)/(sqrt(x+1)-1)# as #x->0# using l'Hospital's Rule or otherwise?

In our case

Using L'Hôpital's Rule, differentiate the numerator and denominator separately with respect to ( x ) and then evaluate the limit again.

First, differentiate the numerator: [ \lim_{x \to 0} \frac{d}{dx}(\sqrt{x^2+1}-1) = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{x^2+1}}\cdot 2x}{\frac{1}{2\sqrt{x+1}}} ]

Next, differentiate the denominator: [ \lim_{x \to 0} \frac{d}{dx}(\sqrt{x+1}-1) = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{x+1}}}{\frac{1}{2\sqrt{x+1}}} ]

Now, evaluate the limit again: [ \lim_{x \to 0} \frac{\frac{1}{2\sqrt{x^2+1}}\cdot 2x}{\frac{1}{2\sqrt{x+1}}} = \lim_{x \to 0} \frac{x}{\sqrt{x^2+1}} ]

Since this is an indeterminate form ( \frac{0}{0} ), apply L'Hôpital's Rule once more: [ \lim_{x \to 0} \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+1}}\right) = \lim_{x \to 0} \frac{\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}}{(x^2+1)^{3/2}} ]

Simplify the expression: [ \lim_{x \to 0} \frac{\frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}}{(x^2+1)^{3/2}} = \lim_{x \to 0} \frac{1-x^2}{(x^2+1)^2} ]

Now, substitute ( x = 0 ) into the expression: [ \lim_{x \to 0} \frac{1-x^2}{(x^2+1)^2} = \frac{1-0^2}{(0^2+1)^2} = \frac{1}{1} = 1 ]

Therefore, ( \lim_{x \to 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x+1}-1} = 1 ).