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How can I determine limit:

$$\lim_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$$

Thanks

user73276
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5 Answers5

9

If you are unfamiliar with L'Hopital's rule, this is what you can do.

$$ \lim_{x \rightarrow 4} \frac{ x - 4 } { x - \sqrt{x} - 2} = \lim_{x \rightarrow 4} \frac{ (\sqrt{x} - 2) ( \sqrt{x} + 2) } { (\sqrt{x} -2) ( \sqrt{x} +1) }= \lim_{x \rightarrow 4} \frac{ ( \sqrt{x} + 2) } { ( \sqrt{x} +1) } = \frac{ 4}{3}.$$

Calvin Lin
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Note that the denominator tends to $4-\sqrt{4-2} = 4-\sqrt2$ as $x \to 4$ and the numerator tends to $0$ as $x \to 4$.

EDIT

The answer to the new question is as follows. We have $$\dfrac{x-4}{x-\sqrt{x}-2} = \dfrac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)} = \dfrac{2+\sqrt{x}}{1+\sqrt{x}} \,\,\,\,\, \forall x \neq 4$$ Now take the limit as $x \to 4$.

2

Write $t=\sqrt x$, now we can write the limit as:

$$\lim_{t\to 2}\frac{t^2-4}{t^2-t-2}=\lim_{t\to2}\frac{(t-2)(t+2)}{(t-2)(t+1)}=\lim_{t\to 2}\frac{t+2}{t+1}=\frac{2+2}{2+1}=\frac43$$

Asaf Karagila
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Let $h=x-4$ then $$\frac{x-4}{x-\sqrt[]{x}-2}=\frac{h}{h-\sqrt[]{h+4}+2}=\frac{h}{h-2\sqrt[]{1+h/4}+2}\sim_0\frac{h}{h-2(1+h/8)+2}=\frac{4}{3}$$

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The post has been edited, now the solution is easy if you apply L'Hopital rule: $$\lim_{x \to 4} \frac{x - 4}{x - \sqrt{x} - 2} = \lim_{x \to 4}\frac{1}{1 - \frac{1}{2\sqrt{x}}} = \frac{1}{1 - \frac{1}{2\sqrt{4}}} = \frac{4}{3}$$

  • It looks like the expression has settled down, so this is a good place to use LHR (I tried looking for a "conjugate factor" approach, but I don't see a tidy one...). – colormegone Apr 18 '13 at 21:34
  • @RecklessReckoner You should be using factorization instead. Works much better, and is the approach most students are familiar with. – Calvin Lin Apr 18 '13 at 21:36
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    Actually, in my experience at this point in the semester, most students would use the "hammer" of l'Hopital... – colormegone Apr 18 '13 at 21:38
  • I think so too... –  Apr 18 '13 at 21:41