Homework 6, due July 26 at beginning of class.
- Find the oblique asymptotes (don't graph anything, just solve for the lines, you also don't have to compute any limits, but show how you solved for the line):
- $\frac{x^{2}+4x+10}{x+2}$
Answer: We do polynomial long division to show that $x^{2}+4x+10=(x+2)(x+2)+6$, and so
\[\frac{x^{2}+4x+10}{x+2}=\frac{(x+2)(x+2)+6}{x+2}=x+2+\frac{6}{x+2},\]
and hence the oblique asymptote is $x+2$.
- $\frac{x^{3}+3x^{2}+4}{x^{2}+2x+1}$
Answer:
Again, long division shows that the numerator is equal to $(x^{2}+2x+1)(x+1)-3x+3$, and so
\[ \frac{x^{3}+3x^{2}+4}{x^{2}+2x+1}=\frac{(x^{2}+2x+1)(x+1)-3x+3}{x^{2}+2x+1}=x+1+\frac{-3x+3}{x^{2}+2x+1},\]
and so $x+1$ is the oblique asymptote.
- $\frac{3x^{4}-6x^{3}+1}{x^{3}-2x^{2}+1}$.
- Compute the following limits using l'Hospitals rule
- $\lim_{x\rightarrow 0} \frac{\sin x}{x}$
Since the limit of the top and bottom of this fraction as $x\rightarrow 0$ are both zero, we can apply l'Hospital's rule,
\[\lim_{x\rightarrow 0}\frac{\sin x}{x}=\lim_{x\rightarrow 0}\frac{(\sin x)'}{(x)'}=\lim_{x\rightarrow 0}\frac{\cos x}{1}=\frac{\cos 0}{1}=1.\]
- $\lim_{x\rightarrow 0} \frac{\cos x}{x^{2}}$ This is a typo, you can't use l'Hospital's rule on this one, it will not be graded.
- $\lim_{x\rightarrow 0}\frac{\sin (\sin x)}{x}$
Again, the limit of the top and bottom are both zero, so we use l'Hospitals rule,
\lim_{x\rightarrow 0}\frac{\sin (\sin x)}{x} =\lim_{x\rightarrow 0}\frac{(\sin (\sin x))'}{(x)'}=\lim_{x\rightarrow 0}\frac{\cos\sin x(\cos x)}{1}=\frac{(\cos\sin 0)(\cos 0)}{1}=1.\]
- $\lim_{x\rightarrow 1} \frac{e^{2x}-e^{2}}{e^{x}-e}$
The limit of the top and bottom are both zero, so l'Hopital's rule:
\[\lim_{x\rightarrow 1} \frac{e^{2x}-e^{2}}{e^{x}-e}=\lim_{x\rightarrow 1} \frac{(e^{2x}-e^{2})'}{(e^{x}-e)'}=\lim_{x\rightarrow 1}\frac{2e^{2x}}{e^{x}}=\frac{2e^{2}}{e^{1}}=2e.\]
- $\lim_{x\rightarrow\infty} (\sin e^{-x})^{\frac{1}{x}}$
For this, we take $e^{\ln}$ of this to get
\lim_{x\rightarrow\infty} (\sin e^{-x})^{\frac{1}{x}}=\lim_{x\rightarrow\infty} e^{\ln (\sin e^{-x})^{\frac{1}{x}}}
=\lim_{x\rightarrow\infty} e^{\frac{1}{x}\ln (\sin e^{-x})}.\]
So we just need to compute the limit of the exponent. It is a product of two functions: $\frac{1}{x}$ (which goes to zero in the limit) and $\ln \sin e^{-x}$ (which goes to $-\infty$, since $e^{-x}\rightarrow 0$, hence so does $\sin e^{-x}$, and $\lim_{x\rightarrow\infty} \ln \sin e^{-x} \lim_{y\rightarrow 0} \ln y=-\infty$). Hence, the limit is inteterminate. For such an inteterminate limit, we divide one of the functions by one over the other, and then use l'Hospital's rule, as follows:
\[\lim_{x\rightarrow \infty} \frac{1}{x}\ln \sin e^{-x}=\lim_{x\rightarrow\infty}\frac{\ln \sin e^{-x}}{1/(1/x)}=\lim_{x\rightarrow\infty}\frac{\ln \sin e^{-x}}{x}\]
and then we use l'Hospitals rule by differentiating the top and bottom, so thisis equal to
\[\lim_{x\rightarrow\infty} \frac{\frac{1}{\sin e^{-x}} (\cos e^{-x})(-e^{-x}) }{1} = \lim_{x\rightarrow \infty} -\frac{e^{-x}\cos e^{-x}}{\sin e^{-x}}.\]
Note that $\lim_{x\rightarrow\infty} \cos e^{-x}=\cos (\lim_{x\rightarrow 0} e^{-x})=\cos 0=1,$ so we only need to compute the limit of $-\frac{e^{-x}}{\sin e^{-x}}$. Again, the top and bottom go to zero, hence we use l'Hospital's rule once more to get
\[\lim_{x\rightarrow \infty} -\frac{e^{-x}}{\sin e^{-x}}=\lim_{x\rightarrow \infty} -\frac{-e^{-x}}{-e^{-x}\cos e^{-x}}=\lim_{x\rightarrow\infty} -\frac{1}{\cos e^{-x}}=-1,\]
and hence
\[\lim_{x\rightarrow\infty} -\frac{e^{-x}\cos e^{-x}}{\sin e^{-x}} = (\lim_{x\rightarrow\infty} \frac{1}{\cos e^{-x}})(\lim_{x\rightarrow \infty} -\frac{e^{-x}}{\sin e^{-x}})=1\cdot(-1)=-1,\]
and thus \[\lim_{x\rightarrow\infty} (\sin e^{-x})^{\frac{1}{x}}=e^{-1}=\frac{1}{e}\]
- $\lim_{x\rightarrow\infty} (1+x)^{\frac{1}{x}}$
Again, we take $e^{\ln}$ to get
\[(1+x)^{\frac{1}{x}}=e^{\ln (1+x)^{\frac{1}{x}}}=e^{\frac{1}{x}\ln (1+x)}=e^{\frac{\ln(1+x)}{x}}\],
so to compute our limit, we just need to compute the limit of the exponent, i.e. the limit of $\frac{\ln(1+x)}{x}$. As $x\rightarrow\infty$, both the top and bottom go to infinity, so we may use l'Hospital's rule:
\[\lim_{x\rightarrow\infty}\frac{\ln(1+x)}{x}=\lim_{x\rightarrow\infty}\frac{1/(1+x)}{1}=0,\]
and hence
\[\lim_{x\rightarrow\infty}(1+x)^{\frac{1}{x}}=\lim_{x\rightarrow\infty} e^{\ln (1+x)^{\frac{1}{x}}}=e^{\lim_{x\rightarrow\infty} \ln (1+x)^{\frac{1}{x}}}=e^{0}=1.\]
- $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}$.
The first few steps to this problem are exactly the same as the previous, only now we need to compute $\lim_{x\rightarrow 0}\frac{\ln(1+x)}{x}$ as $x\rightarrow 0 $ (as opposed to $x\rightarrow\infty$). Since the limits of the top and bottom are both zero (recall that $\ln(1)=0$), we may use l'Hospital's rule to compute
\[\lim_{x\rightarrow 0}\frac{\ln(1+x)}{x}=\lim_{x\rightarrow 0} \frac{1/(1+x)}{1}=1,\]
and thus
\[\lim_{x\rightarrow 0} (1+x)^{\frac{1}{x}}=e^{\lim_{x\rightarrow 0} \frac{\ln(1+x)}{x}}=e^{1}=e.\]
- $\lim_{x\rightarrow\infty} (\ln x)\sin \frac{1}{\ln x}$
This limit is inteterminate since it is the limit of a product of a function going to $\infty $, $\ln x$, and a function going to zero, $\sin \frac{1}{\ln x}$. To see this latter fact, note that $\lim_{x\rightarrow\infty} \ln x=\infty$, hence $\lim_{x\rightarrow \infty}\frac{1}{\ln x}=0$, so $\lim_{x\rightarrow\infty}\sin \frac{1}{\ln x}=\sin \lim_{x\rightarrow\infty} \frac{1}{\ln x}=\sin 0=0$. Hence, we rewrite this as
\[\frac{\sin \frac{1}{\ln x}}{\frac{1}{\ln x}}\]
and comptue the limit of this using l'Hospital's rule (since now both top and bottom are converging to zero), hence
\[\lim_{x\rightarrow\infty} \frac{\sin \frac{1}{\ln x}}{\frac{1}{\ln x}}
=\lim_{x\rightarrow \infty} \frac{(\sin \frac{1}{\ln x})'}{(\frac{1}{\ln x})'}
=\lim_{x\rightarrow\infty} \frac{ (\cos \frac{1}{\ln x})(\frac{1}{\ln x})'}{(\frac{1}{\ln x})'}=\lim_{x\rightarrow\infty} \cos\frac{1}{\ln x}=\cos 0=1.\]