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Find $ \lim\limits_{ x\to 100 } \dfrac { 10-\sqrt { x } }{ x-100 }$

(without using a calculator and other machines...?)

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4  
Can we use the machine inside our head? –  user17762 Jul 2 '12 at 0:15

7 Answers 7

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,

$$ \lim_{x \to 100} \frac{10 - \sqrt{x}}{x - 100} = \lim_{x \to 100} \frac{-\frac{1}{20} (x - 100) - r(x) (x - 100)}{x - 100} = \lim_{x \to 100} -\frac{1}{20} - r(x) = -\frac{1}{20}$$

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$\lim\limits_{ x\to 100 } \dfrac { 10-\sqrt { x } }{ x-100 }=\lim\limits_{ x\to 100 } \dfrac {-(\sqrt { x }-10) }{ (\sqrt{x}-10)(\sqrt{x}+10) }=-\lim\limits_{ x\to 100 } \dfrac {1}{\sqrt{x}+10}=-\frac{1}{20}$

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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}} $

$=- \frac{1}{20}$

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If you are going to recognize it as a derivative, then your second line should not have $\lim$ at the beginning. –  Arturo Magidin Jul 3 '12 at 5:57
    
thank you...it's my mistake –  Siddhant Trivedi Jul 3 '12 at 6:09

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 $\dfrac{-1}{20}$.

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If you are going to use derivatives, it might be worth noting that the quotient has the form $\dfrac{f(x)-f(100)}{x-100}$ (with $f(x)=-\sqrt x$). Generally $f'(c)=\lim\limits_{x\to c}\dfrac{f(x)-f(c)}{x-c}$ by definition. One could apply l'Hôpital's rule, yielding $f'(c) = \lim\limits_{x\to c}f'(x)$ if the latter limit exists. Incidentally, this shows that derivatives cannot have removable discontinuities. –  Jonas Meyer Jul 3 '12 at 6:10

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.

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Multiply the whole expression by $$\frac{10+\sqrt{x}}{10+\sqrt x}\Longrightarrow \frac{100-x}{(x-100)(10+\sqrt x)}=-\frac{1}{10+\sqrt x}\xrightarrow [x\to 100]{}-\frac{1}{20}$$

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Thanks so much!! –  Jiwon Jul 2 '12 at 0:17

Hint: $a^2 - b^2 = (a-b)(a+b)$

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