How to find $ \lim\limits_{ x\to 100 } \frac { 10-\sqrt { x } }{ x-100 }$ Find $ \lim\limits_{ x\to 100 } \dfrac { 10-\sqrt { x } }{ x-100 }$
(without using a calculator and other machines...?)
 A: Hint: $a^2 - b^2 = (a-b)(a+b)$
A: Multiply the whole expression by
$$\frac{10+\sqrt{x}}{10+\sqrt x}\Longrightarrow \frac{100-x}{(x-100)(10+\sqrt x)}=-\frac{1}{10+\sqrt x}\xrightarrow [x\to 100]{}-\frac{1}{20}$$
A: Would the answer not be perfectly clear if we were dealing with $\dfrac{10-u}{u^2-100}$?
Well, let $u=\sqrt{x}$. Now we are.
A: I'll throw out a L'Hopital approach since it has not been suggested yet. It is applicable because we have a $0/0$ indeterminate case.
$$\lim_{x\rightarrow 100}{\dfrac {10-\sqrt {x}}{x-100}}$$
Note that the derivative of $10-\sqrt{x}$ is simply $\frac{-1}{2\sqrt{x}}$ and the derivative of $x-100$ is just $1.$ So, we have:
$$\lim_{x\rightarrow 100}{\dfrac {-1}{2\sqrt{x}}}$$ 
Plug in $x=100$ and you find that the limit is indeed $\dfrac{-1}{20}$.
A: first let minus outside then we get,
$ =\displaystyle \lim_{x\to 100} - \frac{\sqrt{x}-\sqrt{100}}{x-100}$
=$ -\frac{1}{2} (100)^{-\frac{1}{2}} $ $ (\because \displaystyle \lim_{x\to a} \frac{x^n-a^n}{x-a}=na^{n-1}) $
$=-\frac{1}{2}\frac{1}{\sqrt{100}} $
$=- \frac{1}{20}$
A: $\lim\limits_{ x\to 100 } \dfrac { 10-\sqrt { x } }{ x-100 }=\lim\limits_{ x\to 100 } \dfrac {-(\sqrt { x }-10) }{ (\sqrt{x}-10)(\sqrt{x}+10) }=-\lim\limits_{ x\to 100 } \dfrac {1}{\sqrt{x}+10}=-\frac{1}{20}$
A: Recognizing that $\sqrt{x} \approx 10$ makes the numerator vanish, we may be inspired to use a differential approximation:
$$ \sqrt{x} = 10 + \frac{1}{20}(x - 100) + r(x) (x - 100) $$
where $r(x)$ has the property that $\lim_{x \to 100} r(x) = 0$. Therefore,
$$ \lim_{x \to 100} \frac{10 - \sqrt{x}}{x - 100}
= \lim_{x \to 100} \frac{-\frac{1}{20} (x - 100) - r(x) (x - 100)}{x - 100}
= \lim_{x \to 100} -\frac{1}{20} - r(x) = -\frac{1}{20}$$
