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I am hung up on this limit: $\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$

I must be missing something related to dealing with square roots but I can not for the life of me figure out what.

Here is my work so far:

$\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x} = \lim_{x\to0} \frac{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}$

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

$= \displaystyle \lim_{x\to0} \frac{1+x - 1-x}{\sqrt{1+x} + \sqrt{1-x}}= 0$.

After this I end up with the answer 0, but I know that it should come out to 1. If someone could look over this and see where I am going wrong and point me in the right direction I will be eternally thankful!

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    $\begingroup$ The limit of the numerator is $2$, the limit of the denominator is $0$. So... $\endgroup$ – Julien Jan 20 '13 at 17:08
  • $\begingroup$ Check the numerator of your last step. It should be $(1+x)-(1-x) = 2x$. Check the denominator of your last step. It seems like you're missing an $x$ from the previous step. $\endgroup$ – Calvin Lin Jan 20 '13 at 17:13
  • $\begingroup$ There's no need to be fancy. Look at the original limit and think about julien's comment. You should also consider the left and right hand limits separately. $\endgroup$ – David Mitra Jan 20 '13 at 17:15
  • $\begingroup$ Thank you, I am trying to work it out and see if I understand the suggestions now. $\endgroup$ – user59012 Jan 20 '13 at 17:22
  • $\begingroup$ Your edit (please when editing the original post, indicate that you have done so, so that answers addressing your initial work do not become irrelevant): you should end with numerator $1 + x - (1 - x) = 1 + x - 1 + x = 2x$ And your denominator: the sqrt. expressions should be subtracted, not added. $\endgroup$ – Namaste Jan 20 '13 at 17:24
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$$\lim\limits_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$$

Note that evaluating $\quad \displaystyle \lim \frac{\sqrt{1+x} + \sqrt{1-x}}{x}\;$ as $\;x\to 0\;$ gives you $\;2\;$ in the numerator, and $\;0\;$ in the denominator. So the task that ultimately remains is to evaluate the limits as $\;x \to 0^+\,$ and as $\;x \to 0^-$.


Hint: (prior to post's edit): Regarding your algebraic manipulations, if you are attempting to "simplify" the expression to make the limit more evident: try multiplying numerator and denominator by $\;\sqrt{1+x} - \sqrt{1-x}$, and be careful with algebra!

$$\lim\limits_{x\to0} \frac{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{(\sqrt{1+x} - \sqrt{1-x})x}$$ $$= \lim_{x \to 0} \frac{2x}{(\sqrt{1+x} - \sqrt{1-x})x}$$ $$=\lim_{x \to 0} \frac{2}{\sqrt{1+x} - \sqrt{1-x}}$$

Now be sure to take the limit as $x \to 0^+$ and as $x\to 0^-$

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  • $\begingroup$ Woops, I ment to type a subtraction in my second step. I believe my third step reflects the outcome with the sign corrected. Thank you for catching that. $\endgroup$ – user59012 Jan 20 '13 at 17:09
  • $\begingroup$ This is going to be a very stupid question I am certain, but how do you get from (sqrt{1+x}+\sqrt{1-x})(sqrt{1+x}-\sqrt{1-x}) = 2x ? $\endgroup$ – user59012 Jan 20 '13 at 17:28
  • $\begingroup$ See my comment below your answer, after your edit. $\endgroup$ – Namaste Jan 20 '13 at 17:29
  • $\begingroup$ Ha, thank you! you are a saint. $\endgroup$ – user59012 Jan 20 '13 at 17:30
  • $\begingroup$ You're welcome, user59012! $\endgroup$ – Namaste Jan 20 '13 at 17:45
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I think that the question you meant to ask was to find $$\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}.$$ This can be seen to be $1$ by binomial expansion of the terms in the numerator, and some cancellation. The limit of the expression you gave doesn't exist.

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