# Need to find $\lim_{x\to4}\bigl((1/\sqrt x)-(1/2)\bigr)/(x-4)$

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

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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

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

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Hint :

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

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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.

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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*}

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

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

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

From here for the given limits have:

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

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