170- Quiz 6

Quiz 6 is here.

Solutions follow.

Problem 1 defines a function by f(x) = \ln(x^3 + 2), and asks to compute f'(e^{1/3} ).

First, we find f'(x). For this, we use the chain rule, recalling that (\ln x)'=1/x.

We have \displaystyle f'(x)=\frac{3x^2}{x^3+2}. Thus \displaystyle f'(e^{1/3})=\frac{3e^{2/3}}{e+2}.

Problem 2 asks to compute \displaystyle \lim_{x\to\infty}\frac{\ln x}{\sqrt x}.

Note that \lim_{x\to\infty}\ln x=\infty=\lim_{x\to\infty}\sqrt x, so we can try to use L’Hôpital’s rule to compute the required limit: L’Hôpital’s rule tells us that if a is a number or \infty, and

  1. \lim_{x\to a}f(x)=\infty=\lim_{x\to a}g(x), and
  2. \displaystyle \lim_{x\to a}\frac{f'(x)}{g'(x)}=A exists,

then

\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=A

as well. (There is a similar version when

\lim_{x\to a}f(x)=0=\lim_{x\to a}g(x),

but we do not need it here.)

In our case, condition 1 holds. As for condition 2, we see that

\displaystyle \frac{(\ln x)'}{(\sqrt x)'}=\frac{\frac1x}{\frac 12 x^{-1/2}}=\frac{2\sqrt x}{x}=\frac 2{\sqrt x},

and therefore \displaystyle \lim_{x\to\infty}\frac{(\ln x)'}{(\sqrt x)'}=\lim_{x\to\infty}\frac 2{\sqrt x}=0.

It follows that

\displaystyle \lim_{x\to\infty}\frac{\ln x}{\sqrt x}=0

as well.

The graph of (\ln x)/\sqrt x, shown below, seems to confirm our computations.

Click the image above to enlarge.

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This entry was posted on Saturday, October 16th, 2010 at 3:45 pm and is filed under 170: Calculus I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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