Limits via asymptotic expansions
Exercises · Derivatives · with worked solutions ·
Exercise 1
Compute the following limits:
\[
\lim_{x\to0}\frac{6(x-\sin x)-x^{3}}{\sin x^{5}}
\]
\[
\lim_{x\to0}\frac{\sinh^{2}x+2(1-\cosh x)}{(1-\cos x)^{2}}
\]
\[
\lim_{x\to0}\frac{4(1-\cos x)^{2}-x^{2}\sin^{2}x}{\sin^{2}x\log(1+x^{4})}
\]
\[
\lim_{x\to0}\frac{1+x\sin x -e^{x^2}}{x\sin(x^3)}
\]
\[
\lim_{x\to0}\frac{x^{2}\cos x-\sinh x^{2}+\frac{1}{2}x^{4}}{x^{2}-\arctan x^{2}}
\]
\[
\lim_{x\to0}\frac{x\sinh x-2\cosh x +2}{(e^{\sin x}-1)^{2}}
\]
\[
\lim_{x\to0}\frac{8\sqrt{1+\sin x}-8-4x+x^{2}}{(2e^{x}-2-2x-x^{2})\cosh^{2}x}
\]
\[
\lim_{x\to0}\frac{2\log(1+\sin x)-2x+x^{2}}{(2e^{x}-2-2x-x^{2})\cos^{2}x}
\]
\[
\lim_{x\to+\infty}x\left(\sqrt{4+\frac{5}{x}}-2\cos \frac{1}{x}\right)
\]
Solution
We determine the principal part of the Maclaurin expansion of the numerator:
\[
6(x-\sin x)-x^{3}=6\left(\frac{x^{3}}{6}-\frac{x^{5}}{120}+o(x^{5})\right)-x^{3}=-\frac{x^{5}}{20}+o(x^{5}).
\]
The denominator is equivalent to \(x^{5}\) :
\[
\sin x^{5}\sim x^{5}.
\]
Hence the given limit is
\[
\lim_{x\to0}\frac{6(x-\sin x)-x^{3}}{\sin x^{5}}=\lim_{x\to0}\frac{-\frac{x^{5}}{20}}{ x^{5}}=-\frac{1}{20}
\]
Solution
We determine the principal part of the Maclaurin expansion of the numerator:
\[
\sinh^{2}x+2(1-\cosh x)=\left(x+\frac{x^{3}}{6}+o(x^{3})\right)^{2}-x^{2}-\frac{x^{4}}{12}+o(x^{4})=\frac{x^{4}}{4}+o(x^{4}).
\]
The denominator is equivalent to \(x^{4}/4\) :
\[
(1-\cos x)^{2}\sim\frac{x^{4}}{4}.
\]
Hence the given limit is
\[
\lim_{x\to0}\frac{\sinh^{2}x+2(1-\cosh x)}{(1-\cos x)^{2}}=\lim_{x\to0}\frac{\frac{x^{4}}{4}}{\frac{x^{4}}{4}}=1.
\]
Solution
We determine the principal part of the Maclaurin expansion of the numerator:
\[
\begin{array}{l}
\ds4(1-\cos x)^{2}-x^{2}\sin^{2}x=4\left(\frac{x^{2}}{2}-\frac{x^{4}}{24}+o(x^{4})\right)^{2}-x^{2}\left(x-\frac{x^{3}}{6}+o(x^{3})\right)^{2}\\
\\
\ds=\frac{x^{6}}{6}+o(x^{6}).
\end{array}
\]
The denominator is equivalent to \(x^{6}\) :
\[
\sin^{2}x\log(1+x^{4})\sim x^{2}\cdot x^{4}=x^{6}.
\]
Hence the given limit is
\[
\lim_{x\to0}\frac{4(1-\cos x)^{2}-x^{2}\sin^{2}x}{\sin^{2}x\log(1+x^{4})}=\lim_{x\to0}\frac{\frac{x^{6}}{6}}{x^{6}}=\frac{1}{6}.
\]
Solution
We determine the principal part of the Maclaurin expansion of the numerator:
\[
\begin{array}{l}
\ds1+x\sin x -e^{x^2}=1+x\left(x-\frac{x^{3}}{6}+o(x^{3})\right)-\left(1+x^2+\frac{x^4}{2}+o(x^4)\right)\\
\\
\ds=-\frac{2}{3}x^4+o(x^{4}).
\end{array}
\]
The denominator is equivalent to \(x^{4}\) :
\[
x\sin(x^3)\sim x\cdot x^{3}=x^{4}.
\]
Hence the given limit is
\[
\lim_{x\to0}\frac{1+x\sin x -e^{x^2}}{x\sin(x^3)}=\lim_{x\to0}\frac{-\frac{2}{3}x^4}{x^{4}}=-\frac{2}{3}.
\]
Solution
We determine the principal part of the Maclaurin expansion of the numerator:
\[
\begin{array}{l}
\ds x^{2}\cos x-\sinh x^{2}+\frac{1}{2}x^{4}\\
\\
\ds=x^{2}\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}+o(x^{4})\right)-
\left(x^{2}+\frac{x^{6}}{6}+o(x^{6})\right)+\frac{1}{2}x^{4}=-\frac{x^{6}}{8}+o(x^{6}).
\end{array}
\]
For the denominator, from the expansion \(\arctan x=x-\frac{x^{3}}{3}+o(x^{3})\) , it follows that
\[
x^{2}-\arctan x^{2}=\frac{x^{6}}{3}+o(x^{6}).
\]
Hence the given limit is:
\[
\lim_{x\to0}\frac{x^{2}\cos x-\sinh x^{2}+\frac{1}{2}x^{4}}{x^{2}-\arctan x^{2}}=
\lim_{x\to0}\frac{-\frac{x^{6}}{8}}{\frac{x^{6}}{3}}=-\frac{3}{8}.
\]
Solution
We determine the principal part of the Maclaurin expansion of the numerator:
\[
\begin{array}{l}
\ds x\sinh x-2\cosh x +2\\
\\
\ds=x\left(x+\frac{x^{3}}{6}+o(x^{3})\right)-2\left(1+\frac{x^{2}}{2}+
\frac{x^{4}}{24}+o(x^{4})\right)+2=\frac{x^{4}}{12}+o(x^{4}).
\end{array}
\]
The denominator is equivalent to \(x^{2}\) :
\[
(e^{\sin x}-1)^{2}\sim (\sin x)^{2}\sim x^{2}.
\]
Hence the given limit is
\[
\lim_{x\to0}\frac{x\sinh x-2\cosh x +2}{(e^{\sin x}-1)^{2}}=\lim_{x\to0}\frac{\frac{x^{4}}{12}}{x^{2}}
=\lim_{x\to0}\frac{x^{2}}{12}=0.
\]
Solution
We expand to order \(3\) the composite function \(8\sqrt{1+\sin x}\) with initial point \(x=0\) . From \(\sin x\sim x-x^{3}/6\) and \(8\sqrt{1+y}\sim 8+4y-y^{2}+y^{3}/2\) , by composition, we have
\[
\begin{array}{l}
\ds8\sqrt{1+\sin x}=8+4\left(x-\frac{x^{3}}{6}\right)-\left(x-\frac{x^{3}}{6}\right)^{2}
+\frac{1}{2}\left(x-\frac{x^{3}}{6}\right)^{3}+o(x^{3})\\
\\
\ds=8+4x-x^{2}-\frac{1}{6}x^{3}+o(x^{3}).
\end{array}
\]
It follows that the numerator, in the given limit, is equivalent to \(-x^{3}/6\) .
In the denominator we have the factor \(\cosh^{2}x\) which converges to \(1\) , while
\[
2e^{x}-2-2x-x^{2}\sim x^{3}/3.
\]
The given limit is
\[
\lim_{x\to0}\frac{8\sqrt{1+\sin x}-8-4x+x^{2}}{(2e^{x}-2-2x-x^{2})\cosh^{2}x}=\lim_{x\to0}\frac{-\frac{x^{3}}{6}}{\frac{x^{3}}{3}}=
-\frac{1}{2}.
\]
Solution
We expand to order \(3\) the composite function \(2\log(1+\sin x)\) with initial point \(x=0\) . From \(\sin x\sim x-x^{3}/6\) and \(2\log(1+y)\sim 2y-y^{2}+2y^{3}/3\) , by composition, we have
\[
\begin{array}{l}
\ds2\log(1+\sin x)=2\left(x-\frac{x^{3}}{6}\right)-\left(x-\frac{x^{3}}{6}\right)^{2}
+\frac{2}{3}\left(x-\frac{x^{3}}{6}\right)^{3}+o(x^{3})\\
\\
\ds=2x-x^{2}+\frac{1}{3}x^{3}+o(x^{3}).
\end{array}
\]
It follows that the numerator, in the given limit, is equivalent to \(x^{3}/3\) .
In the denominator we have the factor \(\cos^{2}x\) which converges to \(1\) , while
\[
2e^{x}-2-2x-x^{2}\sim x^{3}/3.
\]
The given limit is
\[
\lim_{x\to0}\frac{2\log(1+\sin x)-2x+x^{2}}{(2e^{x}-2-2x-x^{2})\cos^{2}x}=\lim_{x\to0}\frac{\frac{1}{3}x^{3}}{\frac{1}{3}x^{3}}=1.
\]
Solution
\[
\lim_{x\to+\infty}x\left(\sqrt{4+\frac{5}{x}}-2\cos \frac{1}{x}\right)=\lim_{x\to+\infty}2x\left(\sqrt{1+\frac{5}{4x}}-\cos \frac{1}{x}\right)
\]
From \(\cos y=1+o(y)\) and \(\sqrt{1+y}= 1+\frac{1}{2}y+o(y)\) , by composition, we have
\[\begin{align*}
\lim_{x\to+\infty}2x\left(\sqrt{1+\frac{5}{4x}}-\cos \frac{1}{x}\right)&=\lim_{x\to+\infty}2x\left(1+\frac{1}{2}\cdot\frac{5}{4x}+o\left(\frac{1}{x}\right)-1+o\left(\frac{1}{x}\right)\right) \\
&=\lim_{x\to+\infty}2x\left(\frac{5}{8x}+o\left(\frac{1}{x}\right)\right) \\
&=\frac{5}{4}+o(1) \\
&=\frac{5}{4}
\end{align*}\]