Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

This one is frustrating $$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$$

share|cite|improve this question
Did you try to use anon's hint in your last question for this one? – Eugene Jun 20 '12 at 6:09
I did but I couldn't figure out how to apply it to this one. I wasn't sure if it was the same case – user69 Jun 20 '12 at 6:14
Hint: $x-4=(\sqrt{x}-2)(\sqrt{x}+2)$, & $$\frac{1}{\sqrt{x}}-\frac{1}{2}=\frac{2-\sqrt{x}}{2\sqrt{x}}.$$ – anon Jun 20 '12 at 10:40
up vote 3 down vote accepted

Let $f(x)=\frac{1}{\sqrt{x}}$, then $f$ is differentiable on $]0,+\infty[$ and $f'(x)=\frac{-1}{2x^{\frac{3}{2}}}$ and $\displaystyle\lim_{x\rightarrow 4}\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}=f'(4)=-\frac{1}{16}$

share|cite|improve this answer

Hint :

$$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4} = \lim_{x\to 4}\frac{2-\sqrt x}{2\sqrt x(x-4)}=\lim_{x\to 4}\frac{-1}{2\sqrt x (2+\sqrt x)}.$$

share|cite|improve this answer

Note that $$\frac{1}{\sqrt{x}}-\frac{1}{2}=-\frac{\sqrt{x}-2}{2\sqrt{x}}=-\frac{\sqrt{x}-2}{2\sqrt{x}}\frac{\sqrt{x}+2}{\sqrt{x}+2}=-\frac{x-4}{2x+4\sqrt{x}},$$ so the limit becomes fairly simple to evaluate, after cancellation.

share|cite|improve this answer
Yes I have, but I'm supposed to do it the "long" way – user69 Jun 20 '12 at 6:06
Ah! Well, in that case, let me edit my answer. – Cameron Buie Jun 20 '12 at 6:08
You mean you're supposed to do an $\epsilon-\delta$ proof? – Andrew Salmon Jun 20 '12 at 6:24
Fortunately, if $\varepsilon$-$\delta$ is what is needed, the breakdown given should make that a (relative) cakewalk. – Cameron Buie Jun 20 '12 at 6:26

Using L'Hospital rule, we get \begin{equation*} \begin{split} \lim_{x\to 4}\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4} &= \lim_{x\to 4}\frac{-\frac{1}{2x^{\frac{3}{2}}}}{1}\\ &=-\frac{1}{2({4})^{\frac{3}{2}}}=-\frac{1}{16}. \end{split} \end{equation*}

share|cite|improve this answer

$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$.

We substition $\sqrt {x}=t$, hance we:

If $ x\longrightarrow 4\Rightarrow t\longrightarrow 2. $

From here for the given limits have:

$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$=$\lim\limits_{t\to 2} \frac{\frac{1}{t} - \frac12}{t^2-4}$=$\lim\limits_{t\to 2} \frac{\frac{2-t}{2t}}{t^2-4}$=$\lim\limits_{t\to 2} \frac{2-t}{2t(t-2)(t+2)}$=$-\lim\limits_{t\to 2} \frac{t-2}{2t(t-2)(t+2)}$=$-\lim\limits_{t\to 2} \frac{1}{2t(t+2)}$=$-\frac{1}{16}$

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.