# Evaluating limit [closed]

Evaluate the limit $\lim_{x\to 0} (\frac{\sqrt{x^2+1}-1}{(\sqrt{x^2+16}-4})$ I know the limit is 4 by looking at the graph of the function, but how can I algebraically show that that is the limit?

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## closed as off-topic by Jonas Meyer, Claude Leibovici, Chappers, user91500, wythagorasJun 10 at 13:22

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Just multiply the numerator and the denominator by $$(\sqrt{x^2+1}+1)(\sqrt{x^2+16}+4)$$ to get

$$\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}$$ $$=\frac{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}{(\sqrt{x^2+16}-4)(\sqrt{x^2+16}+4)}\cdot\frac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}$$

$$=\frac{x^2+1-1}{x^2+16-16}\cdot\frac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}$$

and cancel out $x^2$ as $x\ne0$ as $x\to0$

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Or$$\lim_{x\to 0} (\frac{\sqrt{x^2+1}-1}{(\sqrt{x^2+16}-4})=\lim_{x\to 0} \frac{\sqrt{x^2+1}-1}{x^2}\cdot\lim_{x\to 0} \frac{(\frac{x}{4})^2}{\sqrt{(\frac{x}{4})^2+1}-1}\cdot4= \frac{1}{2}\cdot2\cdot4=4$$

We applied $$\lim_{t\to 0}\frac{(1+t)^r-1}{t} =r$$

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