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How do I solve this limit? I'm stuck.

$$\lim_{x\to 0} {\sqrt{2+x^2}-\sqrt{2-x^2}\over x^2}$$

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Multiply top and bottom by $\sqrt{2+x^2}+\sqrt{2-x^2}$. – Hakim Aug 4 '14 at 11:21
up vote 3 down vote accepted

As suggested in the comment from @Hakim, you can move forward by multiplying the numerator and the denominator by $$\sqrt{2+x^2} + \sqrt{2 - x^2}$$ to get a difference of squares in the numerator:

$$\lim_{x\to 0} {\sqrt{2+x^2}-\sqrt{2-x^2}\over x^2}\cdot \underbrace{\frac{\sqrt{2+x^2} + \sqrt{2-x^2}}{\sqrt{2+x^2} + \sqrt{2 - x^2}}}_{\large =\,1} = \lim_{x\to 0} \frac {2+x^2 -(2 - x^2)}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$$

$$= \lim_{x\to 0} \frac{2\require{cancel}\cancel{x^2}}{\cancel{x^2}(\sqrt{2+x^2} + \sqrt{2-x^2})} = \frac 2{2\sqrt 2}= \frac 1{\sqrt 2}$$

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This is the thorough and correct answer. +1 for taking the time to construct the expression and doing the computation step by step. – Alex Barac Aug 4 '14 at 12:30
Thank you a lot – Bas Aug 4 '14 at 12:34
You're welcome a lot! ;-) – amWhy Aug 4 '14 at 12:38

We have

$$\lim_{x\to 0} {\sqrt{2+x^2}-\sqrt{2-x^2}\over x^2}\times\color{red}{\frac{\sqrt{2+x^2}+\sqrt{2-x^2}}{\sqrt{2+x^2}+\sqrt{2-x^2}}}=\lim_{x\to 0}\frac{2x^2}{2\sqrt 2 x^2}=\frac1{\sqrt2}$$

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-1 for doing other people's homework… – miracle173 Aug 4 '14 at 11:38
How did you get the expression after the fist equal sign? $\sqrt{2+x^2} + \sqrt{2-x^2}$ is definetly not equal to $2\sqrt{2}$. Also, if you replace the $x$ inside an expression you would need to do it in the whole expression, not only in one part of it. – Alex Barac Aug 4 '14 at 12:06
You said "$\sqrt{2+x^2}+\sqrt{2-x^2}$ is definitely not equal to $2\sqrt2$" and I say that it's definitely equal to $2\sqrt2$ on the limit :-)@AlexBarac – user63181 Aug 4 '14 at 12:15
On the limit, true; But you have a limit of a bigger expression, so if you replace the $x$, you must replace it in the whole expression, not in just that small part of it. You never compute the limit in just one part of the expression, then use the result to continue your computation. – Alex Barac Aug 4 '14 at 12:24
Are teachers only assigning 1 homework question nowadays? When I was younger they'd assign 50 questions, and getting a complete answer on 1 of them wasn't anything close to doing someone else's homework. – DanielV Aug 4 '14 at 12:53

This is an indeterminate limit: 0/0. Therefore the most general way to solve this is to use L'Hôpital's rule:

  1. Take the ratio of the independent derivatives of the numerator and of the denominator, and evaluate the results in the same limit as before. That is, evaluate $$\lim_{x \rightarrow c} \frac{f(x)}{g(x)}$$ by computing $$\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}.$$

  2. If the answer remains 0/0, repeat by taking the derivatives of numerator . If the answer is a finite number, that number should be the limit of the original problem.

So applying L'Hôpital's rule, we do indeed get $\frac{1}{\sqrt{2}}$ as the final answer, but for an entirely different line of reasoning than the suggestion to "rationalize the numerator."

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L'Hopitals can get messy especially when there are radicals. – Ali Caglayan Aug 4 '14 at 13:06

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