Asymptotes
Exercises · Derivatives · with worked solutions ·
Exercise 1
Determine the asymptotes (if any) of the following function on its natural domain
\[
f(x)=\sqrt{\frac{x^{3}-x}{x+2}}
\]
Solution
The function \(f\) is defined on
\[
(-\infty,-2)\cup[-1,0]\cup[1,+\infty).
\]
Since
\[
\lim_{x\to-2^{-}}f(x)=+\infty
\]
the line \(x=-2\) is a vertical asymptote.
For \(x\to+\infty\) we have
\[
\lim_{x\to+\infty}\frac{f(x)}{x}=\lim_{x\to+\infty}\sqrt{\frac{x^{3}-x}{x^{3}+2x^{2}}}=1
\]
and
\[
\begin{array}{l}\ds\lim_{x\to+\infty}f(x)-x=\lim_{x\to+\infty}\left(\sqrt{\frac{x^{3}-x}{x+2}}-x\right)
\frac{\sqrt{\frac{x^{3}-x}{x+2}}+x}{\sqrt{\frac{x^{3}-x}{x+2}}+x}=\\
\\
\ds\lim_{x\to+\infty}\left(\frac{x^{3}-x}{x+2}-x^{2}\right)\frac{1}{x\left(\sqrt{\frac{x^{3}-x}{x^{3}+2x^{2}}}+1\right)}=\\
\\
\ds\lim_{x\to+\infty}\frac{-2x-1}{x+2}\frac{1}{\sqrt{\frac{x^{3}-x}{x^{3}+2x^{2}}}+1}=-1
\end{array}
\]
hence we have the asymptote
\[
y=x-1,\ \ x\to+\infty.
\]
Solution
For \(x\to-\infty\) we have
\[
\lim_{x\to-\infty}\frac{f(x)}{x}=\lim_{x\to-\infty}-\sqrt{\frac{x^{3}-x}{x^{3}+2x^{2}}}=-1
\]
and
\[
\begin{array}{l}\ds\lim_{x\to-\infty}f(x)+x=\lim_{x\to-\infty}\left(\sqrt{\frac{x^{3}-x}{x+2}}+x\right)
\frac{\sqrt{\frac{x^{3}-x}{x+2}}-x}{\sqrt{\frac{x^{3}-x}{x+2}}-x}=\\
\\
\ds\lim_{x\to-\infty}\left(\frac{x^{3}-x}{x+2}-x^{2}\right)\frac{1}{-x\left(\sqrt{\frac{x^{3}-x}{x^{3}+2x^{2}}}+1\right)}=\\
\\
\ds\lim_{x\to-\infty}\frac{2x+1}{x+2}\frac{1}{\sqrt{\frac{x^{3}-x}{x^{3}+2x^{2}}}+1}=1
\end{array}
\]
hence we have the asymptote
\[
y=-x+1,\ \ x\to-\infty.
\]
Solution
Using the asymptotic expansion
\[
\sqrt{1+y}=1+\frac{1}{2}y+o(y),\ \ y\to0
\]
and taking into account the equivalence
\[
\frac{2x+1}{x^{2}+2x}\sim\frac{2}{x},\ \ \ x\to\pm\infty,
\]
hence
\[
\frac{2x+1}{x^{2}+2x}\sim\frac{2}{x}+o\left(\frac{1}{x}\right),\ \ \ x\to\pm\infty,
\]
we can also proceed as follows:
\[
\begin{array}{l}
\ds f(x)=|x|\sqrt{1-\frac{2x+1}{x^{2}+2x}}=\\
\\
|x|\sqrt{1-(2/x)+o(1/x)}=|x|(1-(1/x)+o(1/x))=\\
\\
|x|-{\rm sgn}(x)+o(1),\ \ \ x\to\pm\infty
\end{array}
\]
hence we recover the oblique asymptotes \(y=-x+1\) for \(x\to-\infty\) and \(y=x-1\) for \(x\to+\infty\) .
Exercise 2
Determine the asymptotes (if any) of the following function on its natural domain
\[
f(x)=\sqrt{9x^{2}+2x+1}
\]
Solution
The function \(f\) is defined on all of \(\R\) .
For \(x\to+\infty\) we have
\[
\lim_{x\to+\infty}\frac{f(x)}{x}=\lim_{x\to+\infty}\sqrt{9+2/x+1/x^{2}}=3
\]
and
\[
\begin{array}{l}\ds\lim_{x\to+\infty}f(x)-3x=\lim_{x\to+\infty}\left(\sqrt{9x^{2}+2x+1}-3x\right)
\frac{\sqrt{9x^{2}+2x+1}+3x}{\sqrt{9x^{2}+2x+1}+3x}=\\
\\
\ds\lim_{x\to+\infty}\frac{2x+1}{3x\left(\sqrt{1+2/(9x)+1/(9x^{2})}+1\right)}=\frac{1}{3}\end{array}
\]
hence we have the asymptote
\[
y=3x+\frac{1}{3},\ \ \ x\to+\infty.
\]
For \(x\to-\infty\) we have
\[
\lim_{x\to-\infty}\frac{f(x)}{x}=\lim_{x\to-\infty}-\sqrt{9+2/x+1/x^{2}}=-3
\]
and
\[
\begin{array}{l}\ds\lim_{x\to-\infty}f(x)+3x=\lim_{x\to-\infty}\left(\sqrt{9x^{2}+2x+1}+3x\right)
\frac{\sqrt{9x^{2}+2x+1}-3x}{\sqrt{9x^{2}+2x+1}-3x}=\\
\\
\ds\lim_{x\to-\infty}\frac{2x+1}{-3x\left(\sqrt{1+2/(9x)+1/(9x^{2})}+1\right)}=-\frac{1}{3}\end{array}
\]
hence we have the asymptote
\[
y=-3x-\frac{1}{3},\ \ \ x\to-\infty.
\]
Solution
Alternatively, we can proceed as follows:
\[
\begin{array}{l}
f(x)=3|x|\sqrt{1+\frac{2}{9x}+\frac{1}{9x^{2}}}=\\
\\
3|x|\sqrt{1+(2/9x)+o(1/x)}=3|x|(1+(1/9x)+o(1/x))=\\
\\
3|x|+(1/3){\rm sgn}(x)+o(1),\ \ \ x\to\pm\infty
\end{array}
\]
hence we recover the oblique asymptotes \(y=-3x-(1/3)\) for \(x\to-\infty\) and \(y=3x+(1/3)\) for \(x\to+\infty\) .
Exercise 3
Determine the asymptotes (if any) of the following function on its natural domain
\[
f(x)=x\left(e^{-\frac{1}{x}}+\sin\frac{1}{x}\right)
\]
Solution
The function \(f\) is defined for \(x\neq0\) . We have
\[
\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}xe^{-\frac{1}{x}}+\lim_{x\to0^{-}}x\sin\frac{1}{x}=-\infty+0=-\infty
\]
where
\[
\lim_{x\to0^{-}}xe^{-\frac{1}{x}}=-\infty
\]
follows from the fact that \(e^{-1/x}\) is an infinite quantity of higher order than \(1/x\) for \(x\to0^{-}\) and that \(xe^{-1/x}<0\) for \(x<0\) . For the limit
\[
\lim_{x\to0^{-}}x\sin\frac{1}{x}=0
\]
we used the fact that the product of an infinitesimal (\(x\) ) and a bounded function (\(\sin(1/x)\) ) is an infinitesimal. The line \(x=0\) is therefore a vertical asymptote.
Solution
For \(x\to\pm\infty\) we have
\[
\lim_{x\to\pm\infty}\frac{f(x)}{x}=\lim_{x\to\pm\infty}e^{-\frac{1}{x}}+\sin\frac{1}{x}=1
\]
and
\[
\begin{array}{l}\ds\lim_{x\to\pm\infty}f(x)-x=\lim_{x\to\pm\infty}\frac{e^{-\frac{1}{x}}-1}{\frac{1}{x}}+
\frac{\sin\frac{1}{x}}{\frac{1}{x}}=\\
\\
\ds\lim_{y\to0^{\pm}}\frac{e^{-y}-1}{y}+\lim_{y\to0^{\pm}}\frac{\sin y}{y}=-1+1=0\end{array}
\]
hence there is the oblique asymptote
\[
y=x
\]
both for \(x\to-\infty\) and for \(x\to+\infty\) .
Solution
The same result can be obtained with the expansions \(e^{y}=1+y+o(y)\) , \(\sin y=y+o(y)\) for \(y\to0\) :
\[
f(x)=x(1-1/x+o(1/x))+x(1/x+o(1/x))=x-1+1+o(1)=x+o(1)
\]
for \(x\to\pm\infty\) , which means exactly that we have the oblique asymptote
\[
y=x
\]
both for \(x\to-\infty\) and for \(x\to+\infty\) .