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$f(2)=4$, $g(2)=9$, $f'(2)=g'(2)$.

$ \displaystyle \lim_{x \to 2} \frac{ \sqrt{f(x)}-2} { \sqrt{g(x)}-2} $.

Why isn't this limit equal to $0$? Since $f$ and $g$ are differentiable at $x=2$, that means they are continuous, so you should be able to evaluate the limit by direct substitution. Thanks!

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    $\begingroup$ Shouldn't it be $\sqrt{\frac{2}{7}}$? $\endgroup$ – John Douma May 10 '13 at 19:21
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    $\begingroup$ Did he mean $\sqrt{f(x)}-2$ instead of $\sqrt{f(x)-2}$?? $\endgroup$ – rschwieb May 10 '13 at 19:27
  • $\begingroup$ @Ovi The edit is according to the brackets you had. See if this is correct please, because it really appears as it should have been "sqrt(f(x)-2)/(sqrt(g(x)-2)" instead of "sqrt(f(x)-2)/(sqrt(g(x))-2)". $\endgroup$ – Jerry May 10 '13 at 19:28
  • $\begingroup$ Sorry I saw this so late, this isin't correct. The 2's are not in the square roots $\endgroup$ – Ovi May 10 '13 at 19:32
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    $\begingroup$ @Ovi Then the answer $3/2$ is wrong. Change your exercise book. ;-) $\endgroup$ – egreg May 10 '13 at 20:21
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The conditions given don't make much sense unless the limit you have to compute is $$ \lim_{x\to 2}\frac{\sqrt{f(x)}-2}{\sqrt{g(x)}-3} $$ where direct substitution is impossible. In this case the data about the derivatives allow you to use L'Hôpital's theorem and do

$$ \lim_{x\to 2}\frac{\sqrt{f(x)}-2}{\sqrt{g(x)}-3} = \frac{\dfrac{f'(2)}{2\sqrt{f(2)}}}{\dfrac{g'(2)}{2\sqrt{g(2)}}} = \frac{f'(2)}{4}\frac{6}{g'(2)} =\frac{3}{2} $$ with the further assumption that $f'(2)=g'(2)\ne0$. If the common value of the derivatives at $2$ is zero you can't say anything about the limit with the available information.

If the limit is really

$$ \lim_{x\to 2}\frac{\sqrt{f(x)}-2}{\sqrt{g(x)}-2} $$

then it is $0$ by substituting.

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  • $\begingroup$ This looks like a very good guess at what is wrong. It is a single typo and makes the derivative information useful. $\endgroup$ – Ross Millikan May 10 '13 at 20:06
  • $\begingroup$ @RossMillikan I assumed the information is there for a purpose and not just for misleading the student. The fact that Ovi confirms that $3/2$ is expected tells me there was a typo somewhere. $\endgroup$ – egreg May 10 '13 at 20:14
  • $\begingroup$ Yea I think the website has a typo, i checked again and it is -2, not -3 $\endgroup$ – Ovi May 10 '13 at 20:19
  • $\begingroup$ One shouldn't trust what some so-called math expert tells you on some website. Even if it is self-referential. $\endgroup$ – Hagen von Eitzen May 10 '13 at 20:21
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Substitution is perfectly appropriate, since the denominator does not evaluate to $0$. But I think you substituted $2 = f(x)$ instead of using $f(2) = 4$, and likewise, you substituted $g(x) =2$ instead of $g(2) = 9$.

$$\lim_{x\to 2} \frac{\sqrt{f(x) - 2}}{\sqrt{g(x) - 2}} = \frac {\sqrt{4 - 2}}{\sqrt{9-2}} = \sqrt{\frac{2}7}$$

ADDED: Even with the correction in formatting, substitution is perfectly appropriate to use, again, since the denominator does not evaluate to /approach $0$:

$$\lim_{x\to 2} \frac{\sqrt{f(x)} - 2}{\sqrt{g(x)} - 2} = \frac {\sqrt{4} - 2}{\sqrt{9} -2} = \frac{0}{1} = 0$$

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  • $\begingroup$ Somebody edited the question wrong, the 2's were not supposed to be included in the square roots. However, the answer was supposed to be 3/2. $\endgroup$ – Ovi May 10 '13 at 19:34
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    $\begingroup$ Yes, Ovi is right. $\endgroup$ – mrs May 10 '13 at 19:34
  • $\begingroup$ I guess the website where I got this from was just wrong then $\endgroup$ – Ovi May 10 '13 at 19:35
  • $\begingroup$ Yes, Ovi...Your answer is correct: the limit IS $0$, so the solution given on the website of $\frac 32$ is clearly wrong. You can be confident that your question was legitimate, though it is always unnerving when what's supposed to be a solution manual is mistaken! $\endgroup$ – Namaste May 10 '13 at 19:42
  • $\begingroup$ Repost of earlier, now deleted comment: There was likely a typo in the problem statement, and not in the solution to what was supposed to be the problem statement: that is, IF the problem statement read $\displaystyle \lim_{x\to 2} \dfrac{\sqrt{f(x)} -2}{\sqrt{g(x)} - \color{blue}{\bf 3}}$, then indeed, we would need to use L'Hopital, and in doing so, would arrive at a limit of $\frac 32$, which would then match the solution. I'd report this to the website where the problem statement appears: then it can be easily corrected, for future users! $\endgroup$ – Namaste May 10 '13 at 22:15
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You can evaluate by direct substitution provided that the function in the denominator does not approach 0 or infinity.

I'm kind of suspecting there is a typo in your OP concerning the square roots, but even if a minor modification is necessary to what is actually there, this should address your question.

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Im not really sure how you are finding 2 f(2)=4 and g(2)=9 so f(2)-2= 4-2 and g(2)=9 2-9=7 so you have $(\frac {2}{7})^{1/2}$ no?

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  • $\begingroup$ Somebody edited the question wrong, the 2's were not supposed to be included in the square roots. However, the answer was supposed to be 3/2. $\endgroup$ – Ovi May 10 '13 at 19:34
  • $\begingroup$ Why did you delete your question "388035/absolute-convergence"? $\endgroup$ – Pedro Tamaroff May 10 '13 at 23:14
  • $\begingroup$ @Faust7 Not sure if this applies to you: please see the example section here $\endgroup$ – rschwieb May 11 '13 at 11:30
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As written, $\displaystyle\lim_{x \to 2}\dfrac{\sqrt{f(x)-2}}{\sqrt{g(x)}-2}=\frac {\sqrt 2}1$ by direct substitution. It would be zero, also by direct substitution, if the numerator were $\sqrt {f(x)}-2$

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  • $\begingroup$ Somebody edited the question wrong, the 2's were not supposed to be included in the square roots. However, the answer was supposed to be 3/2. $\endgroup$ – Ovi May 10 '13 at 19:34
  • $\begingroup$ @Ovi: The original post looks to me like it was sqrt (f(x)-2). Maybe that was a typo. I don't see any way to get $\frac 32$ out of it. $\endgroup$ – Ross Millikan May 10 '13 at 19:38
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    $\begingroup$ Yea if you use L'hopitals rule (which isn't valid here) you get 3/2, I guess the person who made the solutions messed up. $\endgroup$ – Ovi May 10 '13 at 19:40

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