## 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.