Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Compute: $\displaystyle\lim_{x\rightarrow 2}\frac{\sqrt{x+1}-\sqrt{ 1-x}}{x}$

share|cite|improve this question
sqrt x+1 means $\sqrt{x} + 1$ or $\sqrt{x+1}$? What about sqrt 1 - x? I'm guessing $\sqrt{1-x}$ – Pragabhava Oct 25 '12 at 0:07
$(\sqrt3-i)/2$. – Berci Oct 25 '12 at 0:09
L'Hopital is not a good way of learning what the limits are but a really good way to check your answer. – user123454321 Oct 25 '12 at 0:13
I'm not sure that the problem was to compute the limit when x --> 2... – Provost Oct 25 '12 at 0:18

I assume you are interested in $$\lim_{x \to 0} \dfrac{\sqrt{1+x} - \sqrt{1-x}}x$$ Multiply and divide by $\sqrt{1+x} + \sqrt{1-x}$ to get $$\dfrac{\sqrt{1+x} - \sqrt{1-x}}x \times \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \dfrac{(1+x)-(1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \dfrac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}\\ = \dfrac2{\sqrt{1+x} + \sqrt{1-x}}$$ Now can you finish it off?

share|cite|improve this answer

Just evaluate the function at $x = 2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.