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In this question: Solving limit without L'Hôpital

The answer is as follows:

$$\lim_{x\to0} \frac{5-\sqrt{x+25}}{x}=\lim_{x\to0} \frac{(5-\sqrt{x+25)}(5+\sqrt{x+25})}{x(5+\sqrt{x+25})}=\lim_{x\to0} \frac{25-(x+25)}{x(5+\sqrt{x+25})}=-\frac{1}{10}$$

Expanding the fraction makes sense, but I dont understand how we get $ -\frac{1}{10}$ as a result. Because when you put in 0 for x ( which I intuitively did) I get $\frac{0}{0}$ as a result, which doesnt get me anywhere withou l'Hospital.

What step did I miss ?

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  • $\begingroup$ sorry about the duplicate, I would've commented on the original but I dont have enough credit and chat is not very active $\endgroup$ – zython Aug 12 '17 at 12:43
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    $\begingroup$ Hint:$$25-(x+25)=-x$$Now cancel factors. $\endgroup$ – Simply Beautiful Art Aug 12 '17 at 12:45
  • $\begingroup$ Just out of curiosity (because I love words) : does your user name mean something special ? Please answer (I shall tell you why later). $\endgroup$ – Claude Leibovici Aug 12 '17 at 13:08
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    $\begingroup$ zython is the name of the beer ancient Egyptian people used to drink. Look at en.wikipedia.org/wiki/Egyptian_zythos With a friend of mine, we use to say eachother "OK, let's go and have a zython". $\endgroup$ – Claude Leibovici Aug 12 '17 at 13:17
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    $\begingroup$ I was just amazed since I suppose that very few people in the world use (or even know) this word. Cheers. $\endgroup$ – Claude Leibovici Aug 12 '17 at 13:27
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$$\lim_{x\to 0}\frac{25-(x+25)}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-x}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-1}{5+\sqrt{x+25}}$$ Note that after the second step we cancel out the $x$ in the numerator and denominator, and are left with an expression with which we can evaluate at $x=0$.

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  • $\begingroup$ :P Looks good to me. $\endgroup$ – Simply Beautiful Art Aug 12 '17 at 12:48
  • $\begingroup$ ah I see, for a moment I forgot that $-x = -1 * x$ thank you for the answer it is clear to me now $\endgroup$ – zython Aug 12 '17 at 12:50
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Hint:

Let $5-\sqrt{x+25}=y$

$\implies(i)5-y=\sqrt{x+25}\implies x=y^2-10y$

and $(ii)y\to0^-$

Can you take it from here?

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other If $f (x)=\sqrt {x+25} $ then $$\lim_0\frac {f (0)-f (x)}{x}=-f'(0) $$ and $$f'(x)=\frac {1}{2f (x)} $$

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  • $\begingroup$ That's a strange way of writing $\lim_{x\to 0}$ $\endgroup$ – zhw. Aug 13 '17 at 15:13

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