Computing limits of sequences
Exercises · Limits of sequences · with worked solutions ·
1. Computing limits with basic techniques
If:
\[
a_n \rr \ell_a\in \R {\rm ~~~~and~~~~} b_n \rr \ell_b \in \R,
\]
we have:
\[
a_n \pm b_n \rr \ell_a\pm \ell_b, ~~~~~ \frac{a_n}{b_n} \rr \frac{\ell_a}{\ell_b} ~~~~ (b_n, \ell_b\neq 0, {\rm ~eventually}),
\]
\[
a_n \: b_n \rr \ell_a\: \ell_b , ~~~~~ a_n^{b_n} \rr {\ell_a}^{\ell_b} ~~~~ (a_n, \ell_a> 0, {\rm ~eventually}).
\]
If:
\[
a_n \rr \ell_a\in \R, ~~~b_n \rr \ip {\rm ~~~and~~~} c_n \rr \ip,
\]
we have:
\[
a_n + b_n \rr \ell_a \ip = \ip, ~~~a_n - b_n \rr \ell_a \im = \im,
\]
\[
b_n + c_n \rr \ip \ip = \ip, ~~~ - b_n - c_n \rr \im \im = \im.
\]
If:
\[
a_n \rr \ell_a\in \R, ~~~b_n \rr 0 {\rm ~~~and~~~} c_n \rr \infty,
\]
we have:
\[
a_n \:\: c_n \rr \ell_a\:\: \infty = \infty ~~ (\ell_a\neq 0), ~~~~ \frac{a_n}{b_n} \rr \frac{\ell_a}{0} = \infty ~~ (\ell_a\neq 0), ~~~~ \frac{a_n}{c_n} \rr \frac{\ell_a}{\infty} = 0.
\]
The four main indeterminate forms are:
\[
[\ip \im], \quad [0 \cdot \infty], \quad \left[\frac{0}{0}\right] {\rm ~~~~and~~~~} \left[\frac{\infty}{\infty}\right].
\]
There are also three other indeterminate forms derived from \([0 \cdot \infty]\) :
\[
\big[1^{\infty}\big], \quad \big[0^0\big] {\rm ~~~~and~~~~} \big[\infty^0\big].
\]
We have:
\[
\lim_{n \rightarrow +\infty} n^{\alpha} =
\begin{cases}
+\infty & {\rm if~} \alpha >0\\
1 & {\rm if~} \alpha = 0\\
0 & {\rm if~} \alpha < 0
\end{cases}
\qquad
\lim_{n \rightarrow +\infty} a^n =
\begin{cases}
+\infty & {\rm if~} a >1\\
1 & {\rm if~} a = 1\\
0 & {\rm if~} |a| < 1\\
{\rm does~not~exist~} & {\rm if~} a \le -1\\
\end{cases}
\]
Exercise 1
Compute the following limit, which is an (elementary) indeterminate form:
\[
\displaystyle\lim_{n\to+\infty}\frac{n^{2}+2n}{n+1} = \left[\frac{\infty}{\infty}\right]
\]
Solution
\[
\lim_{n\to+\infty}\frac{n^{2}+2n}{n+1}=\lim_{n\to+\infty}\frac{n^{2} \overbrace{\left(1+ \overbrace{\frac{2}{n}}^{\rr 0}\right)}^{\rr 1}}{n \underbrace{\left(1+ \underbrace{\frac{1}{n}}_{\rr 0}\right)}_{\rr 1}}
=\lim_{n\to+\infty}\frac{n^{2}}{n}= \lim_{n\to+\infty} n =+\infty.
\]
Exercise 2
Compute the following limit, which is an (elementary) indeterminate form:
\[
\displaystyle \lim_{n\to+\infty}\frac{n^{4}+5}{n^{5}+7n-1} = \left[\frac{\infty}{\infty}\right]
\]
Solution
\[
\lim_{n\to+\infty}\frac{n^{4}+5}{n^{5}+7n-1}=
\lim_{n\to+\infty}\frac{n^{4}\left(1+\frac{5}{n^{4}}\right)}{n^{5}\left(1+\frac{7}{n^{4}}-\frac{1}{n^{5}}\right)}
=\lim_{n\to+\infty}\frac{n^{4}}{n^{5}}=\lim_{n\to+\infty}\frac{1}{n}=0.
\]
Exercise 3
Compute the following limit, which is an (elementary) indeterminate form:
\[
\displaystyle\lim_{n\to+\infty}\frac{1-n^{2}}{(n+2)^{2}} = \left[\frac{\infty}{\infty}\right]
\]
Solution
\[
\lim_{n\to+\infty}\frac{1-n^{2}}{(n+2)^{2}}=
\lim_{n\to+\infty}\frac{-n^{2}\left(1-\frac{1}{n^{2}}\right)}{n^{2}\left(1+\frac{4}{n}+\frac{4}{n^{2}}\right)}
=\lim_{n\to+\infty}\frac{-n^{2}}{n^{2}}=\lim_{n\to+\infty} - 1=-1.
\]
Exercise 4
Compute the following limit, which is an (elementary) indeterminate form:
\[
\displaystyle\lim_{n\to+\infty}\sqrt{n^{2}+1}-\sqrt{n} = [\ip \im]
\]
Solution
\[
\lim_{n\to+\infty}\sqrt{n^{2}+1}-\sqrt{n}=\lim_{n\to+\infty}n\left(
\underbrace{\sqrt{1+\underbrace{\frac{1}{n^{2}}}_{\rr 0}}}_{\rr 1}
-\underbrace{\sqrt{\frac{1}{n}}}_{\rr 0}\right)=
\lim_{n\to+\infty}n=+\infty.
\]
Solution
Alternative method: multiplying and dividing by \(\sqrt{n^2+1} + \sqrt{n}\) we obtain
\[
\frac{(\sqrt{n^{2}+1}-\sqrt{n})(\sqrt{n^{2}+1}+\sqrt{n})}{\sqrt{n^{2}+1}+\sqrt{n}}=\frac{\left(\sqrt{n^2+1}\right)^2 - \left(\sqrt{n}\right)^2}{\sqrt{n^2+1} + \sqrt{n}} = \frac{n^2 + 1 - n}{\sqrt{n^2+1} + \sqrt{n}},
\]
recalling that \((a-b)(a+b)=a^2-b^2\) . Moreover, we have
\[
\frac{n^2 + 1 - n}{\sqrt{n^2+1} + \sqrt{n}}
= \frac{n^2 \left(1 + \frac{1}{n^2} - \frac{1}{n} \right)}{\sqrt{n^2 \left(1+\frac{1}{n^2}\right)} + \sqrt{n}}
= \frac{n^2 \left(1 + \frac{1}{n^2} - \frac{1}{n} \right)}{n \: \sqrt{ \left(1+\frac{1}{n^2}\right)} + n^{\frac{1}{2}} }
= \frac{n^2 \left(1 + \frac{1}{n^2} - \frac{1}{n} \right)}
{n \: \left( \sqrt{ \left(1+ {\frac{1}{n^2}}\right)} + {\frac{1}{n^{1/2}}} \right)}.
\]
Hence
\[
\lim_{n \rr \ip} \sqrt{n^{2}+1}-\sqrt{n}= \lim_{n\to+\infty} n \:
\frac{ \overbrace{\left(1 + \frac{1}{n^2} - \frac{1}{n} \right)}^{\rr 1}}
{ \underbrace{\left( \sqrt{ \left(1+\frac{1}{n^2}\right)} + {\frac{1}{n^{1/2}}} \right)}_{\rr 1}}=\lim_{n\to+\infty}n=+\infty.
\]
Exercise 5
Compute the following limit, which is an (elementary) indeterminate form:
\[
\displaystyle \lim_{n\to+\infty}e^{n}-2^{n} = [\ip \im]
\]
Solution
\[
\lim_{n\to+\infty}e^{n}-2^{n}= \lim_{n\to+\infty}e^{n} \underbrace{\left(1-\underbrace{\left(\frac{2}{e}\right)^{n}}_{\rr 0}\right)}_{\rr 1}= \lim_{n\to+\infty}e^{n} =+\infty
\]
since
\[
e^{n}\to+\infty {\rm ~~and~~} \left(\frac{2}{e}\right)^{n}\to0.
\]
Exercise 6
Compute the following limit, which is an (elementary) indeterminate form:
\[
\displaystyle \lim_{n\to+\infty}3^{n}+4^{n}-5^{n} = [\ip \ip \im] = [\ip \im]
\]
Solution
\[
\lim_{n\to+\infty}3^{n}+4^{n}-5^{n}=
\lim_{n\to+\infty}-5^{n} \underbrace{\left(1- \underbrace{\left(\frac{3}{5}\right)^{n}}_{\rr 0}- \underbrace{\left(\frac{4}{5}\right)^{n}}_{\rr 0}\right)}_{\rr 1}=\lim_{n\to+\infty}-5^{n}=-\infty
\]
since
\[
-5^{n}\to-\infty, ~~~ \left(\frac{3}{5}\right)^{n}\to 0 {\rm ~~and~~} \left(\frac{4}{5}\right)^{n}\to0.
\]
Exercise 7
Compute the following simple limit
\[
\displaystyle \lim_{n\to+\infty}n^{\sqrt{2}}
\]
Solution
For every \(\alpha>0\) , even irrational, \(\lim_{n\to+\infty}n^{\alpha}=+\infty\) .
Exercise 8
Compute the following simple limit
\[
\displaystyle \lim_{n\to+\infty}n^{-e}
\]
Solution
\[
\lim_{n\to+\infty}n^{-e}=\lim_{n\to+\infty}\frac{1}{n^{e}}=0.
\]
Exercise 9
Compute the following simple limit
\[
\displaystyle \lim_{n\to+\infty} \left( \frac{3 \: n + 5}{ n^2 + 120}\right)^{2\:n}
\]
Solution
We have
\[
\frac{3 \: n + 5}{ n^2 + 120} = \frac{ 3 \: n \left(1 + \frac{5}{3 \: n} \right)}{n^2 \: \left( 1 + \frac{120}{n^2} \right)} = \frac{3}{n} \: \frac{\left(1 + \frac{5}{3 \: n} \right)}{ \left( 1 + \frac{120}{n^2} \right)} \rr 0 {\rm ~~ and ~~} 2\:n \rr \ip.
\]
Hence
\[
\lim_{n\to+\infty} \left( \frac{3 \: n + 5}{ n^2 + 120}\right)^{2\:n} = 0^{\ip} =0.
\]
Exercise 10
Compute the following simple limit
\[
\displaystyle \lim_{n\to+\infty} \left( \frac{10 \: n^2 - 5 \:n}{ n^3 - 8}\right)^{-3\:n + 8}
\]
Solution
We have
\[
\frac{10 \: n^2 - 5 \:n}{ n^3 - 8} = \frac{ 10 \: n^2 \left(1 - \frac{5 \:n}{10 \: n^2} \right)}{n^3 \: \left( 1 - \frac{8}{n^3} \right)} = \frac{10}{n} \: \frac{\left(1 - \frac{1}{2 \: n} \right)}{ \left( 1 - \frac{8}{n^3} \right)} \rr 0 {\rm ~~ and ~~} -3\:n + 8 \rr \im.
\]
Hence
\[
\lim_{n\to+\infty} \left( \frac{10 \: n^2 - 5 \:n}{ n^3 - 8}\right)^{-3\:n + 8} = 0^{\im} = \frac{1}{0^{\ip}} = \frac{1}{0}= \ip.
\]
2. Computing limits using the hierarchy of infinities
For every \(a>1\) and \(\alpha > 0\) we have:
\[
\lim_{n \rightarrow +\infty} \frac{\log_a n}{n^{\alpha}} = 0, \quad
\lim_{n \rightarrow +\infty} \frac{n^{\alpha}}{a^n} = 0,
\]
\[
\lim_{n \rightarrow +\infty} \frac{a^n}{n!} = 0, \quad
\lim_{n \rightarrow +\infty} \frac{n!}{n^n} = 0. \quad
\]
Hence the following infinities are listed in increasing order:
\[
\log n,~~~ n^{\alpha},~~~ a^{n},~~~ n!,~~~ n^{n}.
\]
Exercise 11
Compute the limit of the following sequence, if it exists
\[
a_n = \frac{{2^{1/n}} + n^2 + 3^{-n}}{\log^6 n + 2 + n}
\]
Solution
Consider the numerator; we have:
\[
{2^{1/n}} \rr 1, \quad n^2 \rr \ip, \quad 3^{-n} \rr 0
\]
The power of \(n\) with the highest degree is \(n^2\) , and moreover we have
\[
\frac{3^{-n}}{n^2} = \frac{1}{3^n \; n^2} {\rm ~~~hence~~~} \frac{3^{-n}}{n^2} \rr 0 {\rm ~~~~~and~~}\frac{2^{1/n}}{n^2} \rr 0.
\]
Consider the denominator; we have:
\[
\log^6 n \rr \ip, \quad 2 \rr 2, \quad n \rr \ip
\]
By the hierarchy of infinities theorem, we have
\[
\frac{\log^6 n}{n}= \left(\underbrace{\frac{\log n}{n^{1/6}}}_{\rr 0}\right)^{6} \rr 0.
\]
We therefore factor out \(n^2\) in the numerator and \(n\) in the denominator and obtain:
\[
\frac{n^2 \left(1+ \frac{2^{1/n}}{n^2} + \frac{3^{-n}}{n^2} \right)}{n \left(1+ \frac{\log^6 n}{n} + \frac{2}{n}\right)} = n \; \frac{ \left(1+ \overbrace{\frac{2^{1/n}}{n^2}}^{\rr 0} + \overbrace{\frac{3^{-n}}{n^2}}^{\rr 0} \right)}{\left(1+ \underbrace{\frac{\log^6 n}{n}}_{\rr 0} + \underbrace{\frac{2}{n}}_{\rr 0} \right)}
\]
\[
\lim_{n \rr \ip} \frac{{2^{1/n}} + n^2 + 3^{-n}}{\log^6 n + 2 + n} = \lim_{n \rr \ip } n \; \overbrace{\frac{ \left(1+ \frac{2^{1/n}}{n^2} + \frac{3^{-n}}{n^2} \right)}{\left(1+ \frac{\log^6 n}{n} + \frac{2}{n}\right)}}^{\rr 1} = \ip
\]
Exercise 12
Compute the limit of the following sequence, if it exists
\[
a_n = \frac{ 3 \: n^3 + 3^n + \log n}{\log^6 n + 2^{2\:n} + n^5}
\]
Solution
Consider the numerator; we have:
\[
3 \: n^3 \rr \ip, \quad 3^n \rr \ip, \quad \log n \rr \ip
\]
that is, a sum of sequences tending to infinity. We also have:
\[
\frac{3 \: n^3}{3^n} \rr 0 {\rm ~~~and ~~~} \frac{\log n}{3^n} \rr 0,
\]
by the hierarchy of infinities theorem. Hence the principal part is \(3^n\) . Consider the denominator; we have:
\[
\log^6 n \rr \ip, \quad 2^{2\:n} =4^{n} \rr \ip, \quad n^5 \rr \ip
\]
that is, a sum of sequences tending to infinity. We also have:
\[
\frac{\log^6 n}{4^{n}} \rr 0 {\rm ~~~and ~~~} \frac{n^5}{4^{n}} \rr 0,
\]
by the hierarchy of infinities theorem. Hence the principal part is \(4^n\) .
We therefore factor out \(3^n\) in the numerator and \(4^n\) in the denominator and obtain:
\[
\frac{3^n \left(1+ \frac{3 \: n^3}{3^n} + \frac{\log n}{3^n} \right)}{4^n \left(1+ \frac{\log^6 n}{4^{n}} + \frac{n^5}{4^{n}} \right)} = \left( \frac{3}{4}\right)^n \; \frac{ \left(1+ \overbrace{\frac{3 \: n^3}{3^n}}^{\rr 0} + \overbrace{\frac{\log n}{3^n}}^{\rr 0} \right)}{\left(1+ \underbrace{\frac{\log^6 n}{4^{n}}}_{\rr 0} + \underbrace{\frac{n^5}{4^{n}}}_{\rr 0} \right)}
\]
hence
\[
\lim_{n \rr \ip} \frac{ 3 \: n^3 + 3^n + \log n}{\log^6 n + 2^{2\:n} + n^5} = \lim_{n \rr \ip } \left( \frac{3}{4}\right)^n \; \underbrace{\frac{ \left(1+ \frac{3 \: n^3}{3^n} + \frac{\log n}{3^n} \right)}{\left(1+ \frac{\log^6 n}{4^{n}} + \frac{n^5}{4^{n}} \right)}}_{\rr 1} = 0
\]
Exercise 13
Compute the following simple limit
\[
\displaystyle \lim_{n\to+\infty}\sqrt[n]{n^{2}}
\]
Solution
We have
\[
\sqrt[n]{n} = n^{\frac{1}{n}} = e^{\log n^{1/n}} = e^{ \frac{\log n}{n}}
\]
and, studying the sequence in the exponent,
\[
a_n = \frac{\log n}{n} {\rm ~~~we~have~~~} a_n \rr 0
\]
thanks to the hierarchy of infinities theorem. Hence we have:
\[
\lim_{n \rr \ip} \sqrt[n]{n}= \lim_{n \rr \ip} e^{ \frac{\log n}{n}} =1.
\]
Consequently
\[
\lim_{n\to+\infty}\sqrt[n]{n^{2}}=\lim_{n\to+\infty}\left(\sqrt[n]{n}\right)^{2}=1^{2}=1.
\]
3. Computing limits using asymptotic estimates
We have \(~~ a_n \thicksim b_n~~\) if \(~~ \frac{a_n}{b_n} \rr 1 ~~\) or if \(~~a_n = b_n \: c_n ~~\) with \(~~ c_n \rr 1.\)
Moreover, we have \(~~ a_n + b_n \thicksim a_n ~~\) if \(~~ \frac{b_n}{a_n} \rr 0\)
Exercise 14
Compute the following limit
\[
\lim_{n\to+\infty}\frac{2^{n}+n^{2}+1}{5^{n}+2^{n}+n}
\]
Solution
We have
\[
2^{n}+n^{2}+1 \thicksim 2^{n} {\rm ~~since~~} \frac{n^2}{2^{n}} \rr 0 {~~and~~} \frac{1}{2^{n}} \rr 0
\]
and
\[
5^{n}+2^{n}+n \thicksim 5^{n} {\rm ~~since~~} \frac{2^n}{5^{n}} = \left(\frac{2}{5}\right)^{n} \rr 0 {~~and~~} \frac{n}{5^{n}} \rr 0
\]
hence
\[
\lim_{n\to+\infty}\frac{2^{n}+n^{2}+1}{5^{n}+2^{n}+n}=\lim_{n\to+\infty}\frac{2^{n}}{5^{n}}=
\lim_{n\to+\infty}\left(\frac{2}{5}\right)^{n}=0
\]
Exercise 15
Compute the following limit
\[
\lim_{n\to+\infty}\frac{n!-5^{n}}{7^{n}}
\]
Solution
We have
\[
n!-5^{n} \thicksim n! {\rm ~~since~~} \frac{5^n}{n!} \rr 0
\]
hence
\[
\lim_{n\to+\infty}\frac{n!-5^{n}}{7^{n}}=\lim_{n\to+\infty}\frac{n!}{7^{n}}=+\infty
\]
Exercise 16
Compute the following limit
\[
\lim_{n\to+\infty}\frac{2^{3n-1}-n^{2}}{(2n)!-5^{n}}
\]
Solution
We have
\[
2^{3n-1}-n^{2} \thicksim 2^{3n-1} {\rm ~~since~~} - \frac{n^2}{2^{3n-1}} \rr 0
\]
and
\[
(2n)!-5^{n} \thicksim(2n)! {\rm ~~since~~} -\frac{5^{n}}{(2n)!} \rr 0
\]
hence
\[
\lim_{n\to+\infty}\frac{2^{3n-1}-n^{2}}{(2n)!-5^{n}}=\lim_{n\to+\infty}\frac{2^{3n-1}}{(2n)!}=0
\]
Exercise 17
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{ 3^n}{ n!\: n^{1/n}}
\]
Solution
We have
\[
n!\: n^{1/n} \thicksim n! {\rm ~~since~~} n^{1/n} = \sqrt[n]{n} \rr 1
\]
because
\[
n^{1/n} = e^{\log n^{1/n}} = e^{\frac{\log n}{n}} {\rm ~~ and~~} \frac{\log n}{n} \rr 0.
\]
Hence
\[
\lim_{n \rr \ip} \frac{ 3^n}{ n!\: n^{1/n}} = \lim_{n \rr \ip} \frac{ 3^n}{ n! } = 0.
\]
Exercise 18
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{\log{(1+e^n)}}{\sqrt{1+n^2}}
\]
Solution
Observing that
\[
\log{(1+e^n)} \thicksim \log(e^n)=n\log{e}=n
\]
we have
\[
\lim_{n \rr \ip} \frac{\log{(1+e^n)}}{\sqrt{1+n^2}} = \lim_{n \rr \ip} \frac{n}{\sqrt{1+n^2}} = \lim_{n \rr \ip} \sqrt{\frac{n^2}{1+n^2}} =1
\]
because
\[
\frac{n^2}{1+n^2} \rr 1.
\]
Exercise 19
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{n+1}{\log_2{(3+n^n)}}
\]
Solution
Observing that
\[
\log_2{(3+n^n)} \thicksim \log_2(n^n)=n\log_2{n}
\]
we have
\[
\lim_{n \rr \ip} \frac{n+1}{\log_2{(3+n^n)}} = \lim_{n \rr \ip} \frac{n+1}{n\log_2{n}} = 0
\]
because
\[
\frac{n+1}{n} \rr 1 {\rm ~~~~and~~~~} \frac{1}{\log_2{n}} \rr 0.
\]
4. Computing limits with the sequence converging to Euler's number \(e\)
For indeterminate forms \(1^{\pm\infty}\) , one can use:
\[
\left(1+\frac{1}{a_{n}}\right)^{a_{n}}\to e\ \ \ {\rm as}\ a_{n}\to\pm\infty.
\]
Exercise 20
Compute the following limit
\[
\lim_{n\to+\infty}\left(\frac{n-1}{n-3}\right)^{n}
\]
Solution
We have
\[
\left(\frac{n-1}{n-3}\right)^{n} = \left(\frac{n-3-1+3}{n-3}\right)^{n} = \left( 1 + \frac{2}{n-3}\right)^{n} = \left( 1 + \frac{1}{ \frac{n-3}{2}}\right)^{n}
\]
hence
\[
\lim_{n\to+\infty}\left(\frac{n-1}{n-3}\right)^{n}=
\lim_{n\to+\infty}\left[\left(1+\frac{1}{\frac{n-3}{2}}\right)^{\frac{n-3}{2}}\right]^{\frac{2n}{n-3}}=e^{2},
\]
since
\[
\left(1+\frac{1}{\frac{n-3}{2}}\right)^{\frac{n-3}{2}} \rr e, \quad 2 \: \left( \frac{n}{n-3} \right)= 2 \:\left( \frac{n-3 + 3}{n-3} \right) = 2 \: \left(1 + \underbrace{\frac{3}{n-3}}_{\rr 0} \right) \rr 2
\]
and
\[
\frac{n-3}{2} \rr \ip.
\]
Exercise 21
Compute the following limit
\[
\lim_{n\to+\infty}\left(\frac{n^{2}+1}{n^{2}}\right)^{n}
\]
Solution
\[
\lim_{n\to+\infty}\left(\frac{n^{2}+1}{n^{2}}\right)^{n}=
\lim_{n\to+\infty}\left[\left(1+\frac{1}{n^{2}}\right)^{n^{2}}\right]^{\frac{1}{n}}=e^{0}=1
\]
since
\[
\left(1+\frac{1}{n^2}\right)^{n^2} \rr e, \quad \frac{1}{n} \rr 0
\]
and
\[
n^2 \rr \ip.
\]
Exercise 22
Compute the following limit
\[
\lim_{n\to+\infty}\left(\frac{2n-5}{2n}\right)^{-n}
\]
Solution
\[
\lim_{n\to+\infty}\left(\frac{2n-5}{2n}\right)^{-n}=
\lim_{n\to+\infty}\left[\left(1+\frac{1}{-\frac{2n}{5}}\right)^{-\frac{2n}{5}}\right]^{\frac{5}{2}}=\sqrt{e^{5}}.
\]
since
\[
\left(1+\frac{1}{-\frac{2n}{5}}\right)^{-\frac{2n}{5}} \rr e {\rm ~~~~and~~~~} -\frac{2n}{5} \rr \im.
\]
Exercise 23
Compute the following limit
\[
\lim_{n\to+\infty}\frac{n^{n-1}}{(n-1)^n}
\]
Solution
We have
\[
\frac{n^{n-1}}{(n-1)^n}=
\frac{1}{n}\left(\frac{n}{n-1}\right)^n=\frac{1}{n}\left(\frac{n-1}{n}\right)^{-n}=\frac{1}{n}\left(1+\frac{1}{-n}\right)^{-n}
\]
hence
\[
\lim_{n\to+\infty}\frac{n^{n-1}}{(n-1)^n}=\lim_{n\to+\infty}\frac{1}{n}\left(1+\frac{1}{-n}\right)^{-n}=0
\]
since
\[
\left(1+\frac{1}{-n}\right)^{-n} \rr e {\rm ~~~~and~~~~} \frac{1}{n} \rr 0.
\]
Exercise 24
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \left( \frac{n^2 +2}{n^2 + n +1} \right)^{2\: n}
\]
Solution
Since
\[
\left( \frac{n^2 +2}{n^2 + n +1} \right) \rr 1 {\rm ~~and~~} 2\: n \rr \ip
\]
we have an indeterminate form of type \([1^{+\infty}]\) .
Since
\[
\left( \frac{n^2 +2}{n^2 + n +1} \right)^{2\: n}
=
\left( \frac{n^2 + n +1 - n +1}{n^2 + n +1} \right)^{2\: n}
=
\left( 1 + \frac{- n +1}{n^2 + n +1} \right)^{2\: n},
\]
we can therefore write
\[
\left( 1 + \frac{- n +1}{n^2 + n +1} \right)^{2\: n}
=
\left( 1 + \frac{1}{\frac{n^2 + n +1}{- n +1}} \right)^{2\: n}
=
\left[\left( 1 + \frac{1}{\frac{n^2 + n +1}{- n +1}} \right)^{\frac{n^2 + n +1}{- n +1}}\right]^{ \left( \frac{- n +1}{n^2 + n +1} \right) \: 2\: n}
.
\]
Hence
\[
\lim_{n\to+\infty}\left( \frac{n^2 +2}{n^2 + n +1} \right)^{2\: n}=
\lim_{n\to+\infty}\left[
\left( 1 + \frac{1}{\frac{n^2 + n +1}{- n +1}} \right)^{\frac{n^2 + n +1}{- n +1}}
\right]^{ \left( \frac{- n +1}{n^2 + n +1} \right) \: 2\: n} =e^{-2},
\]
since
\[
\left( 1 + \frac{1}{\frac{n^2 + n +1}{- n +1}} \right)^{\frac{n^2 + n +1}{- n +1}} \rr e, \quad \left( \frac{- n +1}{n^2 + n +1} \right) \: 2\: n = \left( \frac{ \overbrace{- 2 \: n^2 + 2\: n}^{\thicksim -2 \:n ^2}}{ \underbrace{n^2 + n +1}_{\thicksim n ^2}} \right) \rr -2
\]
and
\[
\frac{n^2 + n +1}{- n +1} = \frac{ n^2 \overbrace{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)}^{\rr 1}}{- n \underbrace{\left( 1 - \frac{1}{n}\right)}_{\rr 1}} \rr \im.
\]
5. Computing limits with the comparison theorem
If \(c_n \rr 0\) and \(|b_n|\le c_n\) eventually, then \(b_n \rr 0\) .
If \(c_n \rr 0\) and \(b_n\) is bounded (even if not convergent), then \(c_n \: b_n \rr 0\) . The product of an infinitesimal sequence and a bounded one is infinitesimal.
A sum of infinities and bounded sequences is asymptotically equivalent to the infinity of highest order.
Exercise 25
Compute the following limit
\[
\lim_{n\to+\infty}\frac{n+\sin n}{\cos n+\log n}
\]
Solution
We have
\[
n+\sin n \thicksim n {\rm ~~since~~} \frac{\sin n}{n} \rr 0
\]
the product of an infinitesimal sequence \(\{\frac{1}{n}\}\) and a bounded one \(\{\sin n\}\) . Moreover,
\[
\cos n+\log n\thicksim \log n {\rm ~~since~~} \frac{\cos n}{\log n} \rr 0
\]
the product of an infinitesimal sequence \(\left\{\frac{1}{\log n} \right\}\) and a bounded one \(\{\cos n\}\) . Hence
\[
\lim_{n\to+\infty}\frac{n+\sin n}{\cos n+\log n}=\lim_{n\to+\infty}\frac{n}{\log n}=+\infty
\]
Exercise 26
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{ n \: \sin n + \sin n^2}{ n^2 +1}
\]
Solution
We have
\[
\frac{ n \: \sin n + \sin n^2}{ n^2 +1}
=
\frac{ n \: \sin n }{ n^2 +1} + \frac{ \sin n^2}{ n^2 +1}
\]
and
\[
\frac{ n }{ n^2 +1} \: \sin n \rr 0 {\rm~~since~} \frac{ n }{ n^2 +1} \rr 0 {~~and~~} \sin n {\rm ~~is~bounded.}
\]
Moreover
\[
\frac{ 1 }{ n^2 +1} \: \sin n^2 \rr 0 {\rm~~since~} \frac{ 1 }{ n^2 +1} \rr 0 {~~and~~} \sin n^2 {\rm ~~is~bounded.}
\]
Hence
\[
\lim_{n \rr \ip} \frac{ n \: \sin n + \sin n^2}{ n^2 +1} = \lim_{n \rr \ip}
\underbrace{\frac{ n \: \sin n }{ n^2 +1}}_{\rr 0} + \underbrace{\frac{ \sin n^2}{ n^2 +1}}_{\rr 0} = 0
\]
Solution
Alternative method.
Note that the sequences \(\{\sin n\}\) and \(\{ \sin n^2\}\) are irregular but bounded. Hence we can write the following upper bound
\[
\left | \frac{ n \: \sin n + \sin n^2}{ n^2 +1} \right| \le \frac{n+1}{n^2 + 1}
\]
and we have
\[
\lim_{n \rr \ip} \frac{n+1}{n^2 + 1} = \lim_{n \rr \ip} \frac{n}{n^2} = \lim_{n \rr \ip} \frac{1}{n} = 0.
\]
Hence, by the comparison theorem, we have:
\[
\lim_{n \rr \ip} \frac{ n \: \sin n + \sin n^2}{ n^2 +1} = 0
\]
6. Computing limits with the ratio test
Given a positive sequence (\(a_n > 0\) for every \(n\) ):
\[
{\rm if~there~exists~~} \lim_{n \rr \ip} \frac{a_{n+1}}{a_n}=\ell
{\rm ~~and~~} \ell < 1, {\rm ~~then~~} a_n \rr 0,
\]
\[
{\rm if~there~exists~~} \lim_{n \rr \ip} \frac{a_{n+1}}{a_n}=\ell
{\rm ~~and~~} \ell > 1 {\rm~~(or~~} \ell = \ip), {\rm ~~then~~} a_n \rr \ip.
\]
Exercise 27
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{3^{n-1}}{n\:(n+1)!}
\]
Solution
Since this is a sequence with positive terms whose general term is the ratio of simpler sequences of exponential and factorial type, it is natural to apply the ratio test.
We therefore compute:
\[
\frac{a_{n+1}}{a_n} = \frac{\frac{3^{n}}{(n+1) \cdot (n+2)!}}{\frac{3^{n-1}}{n\:(n+1)!}} = \frac{3^{n}}{ (n+1) \cdot \underbrace{(n+2)!}_{=(n+1)! \cdot (n+2)}} \: \frac{n\:(n+1)!}{3^{n-1}}=
\]
\[
= \frac{3\:n}{(n+1)(n+2)} = \frac{3\:n}{n^2 + 2 \:n + n + 2} = \frac{3\:n}{n^2 \left( 1 + \frac{3}{n} + \frac{2}{n^2}\right)}= \frac{3}{n \left( 1 + \frac{3}{n} + \frac{2}{n^2}\right)}
\]
We have:
\[
\lim_{n \rr \ip} \frac{a_{n+1}}{a_n} =\lim_{n \rr \ip} \frac{3}{n \left( 1 + \frac{3}{n} + \frac{2}{n^2}\right)} =0
\]
Hence, by the ratio test:
\[
\lim_{n \rr \ip} \frac{3^{n-1}}{n\:(n+1)!} = 0
\]
Exercise 28
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{n^3 \: 2^n}{n!}
\]
Solution
We have a positive sequence; we compute:
\[
\frac{a_{n+1}}{a_n} = \frac{(n+1)^3 \: 2^{(n+1)}}{(n+1)!} \frac{n!}{n^3 \: 2^n} = \underbrace{\left(\frac{n+1}{n} \right)^3}_{\rr 1} \frac{2}{n+1} \thicksim \frac{2}{n+1}
\]
We therefore have:
\[
\lim_{n \rr \ip} \frac{a_{n+1}}{a_n} = \lim_{n \rr \ip} \frac{2}{n+1} =0
\]
Then, by the ratio test:
\[
\lim_{n \rr \ip} \frac{n^3 \: 2^n}{n!} = 0
\]
Exercise 29
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{n^n}{(n+1)!}
\]
Solution
We have a positive sequence; we compute:
\[
\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+2)!} \frac{(n+1)!}{n^n} = \underbrace{\frac{n+1}{n+2}}_{\rr 1} \left( \frac{n+1}{n}\right)^n \thicksim \left( 1 + \frac{1}{n}\right)^n
\]
We therefore have:
\[
\lim_{n \rr \ip} \frac{a_{n+1}}{a_n} = \lim_{n \rr \ip} \left( 1 + \frac{1}{n}\right)^n = e > 1
\]
Then, by the ratio test:
\[
\lim_{n \rr \ip} \frac{n^n}{(n+1)!} = \ip
\]
Exercise 30
Compute the limit of the following sequence, if it exists
\[
\lim_{n \rr \ip} \frac{n^{\frac{1}{n}}\: 2^n}{(n+1)!}
\]
Solution
We have:
\[
\frac{ \overbrace{n^{\frac{1}{n}}}^{\rr 1}\: 2^n}{(n+1)!} \thicksim \frac{2^n}{(n+1)!} \equiv b_n
\]
We now have the positive sequence \(\{b_n\}\) ; we compute:
\[
\frac{b_{n+1}}{b_n} = \frac{2^{n+1}}{(n+2)!} \frac{(n+1)!}{2^n} = \frac{2}{n+2}
\]
We therefore have:
\[
\lim_{n \rr \ip} \frac{b_{n+1}}{b_n} = \lim_{n \rr \ip} \frac{2}{n+2} = 0
\]
Then, by the ratio test:
\[
\lim_{n \rr \ip} \frac{n^{\frac{1}{n}}\: 2^n}{(n+1)!} = 0
\]
7. Discussion depending on a parameter
Exercise 31
Compute, as the parameter \(\alpha \in \mathbb{R}\) varies, the limit of the sequence
\[
\lim_{n \rr \ip} \frac{5^{3 \alpha n}}{2^{3n+1}}
\]
Solution
We have
\[
\frac{5^{3 \alpha n}}{2^{3n+1}} = \frac{1}{2}\left(\frac{5^\alpha}{2}\right)^{3n}
\]
hence
\[
\lim_{n \rr \ip} \frac{5^{3 \alpha n}}{2^{3n+1}}=\lim_{n \rr \ip} \frac{1}{2}\left(\frac{5^\alpha}{2}\right)^{3n} = \frac{1}{2} \lim_{n \rr \ip}\left(\frac{5^\alpha}{2}\right)^{3n}=\ell
\]
At this point, we can study three cases:
Case 1
\[
\frac{5^\alpha}{2}>1 \Longleftrightarrow \alpha>\log_5 2
\]
then \(\ell=+\infty\) ;
Case 2
\[
0<\frac{5^\alpha}{2}<1 \Longleftrightarrow \alpha<\log_5 2
\]
then \(\ell=0\) ;
Case 3
\[
\frac{5^\alpha}{2}=1 \Longleftrightarrow \alpha=\log_5 2
\]
then \(\ell=\frac{1}{2}\) .
In conclusion:
\[
\lim_{n \rr \ip} \frac{5^{3 \alpha n}}{2^{3n+1}}=
\begin{cases}
+\infty & \text{if } \alpha > \log_5 2 \\
\frac{1}{2} & \text{if } \alpha = \log_5 2\\
0 & \text{if } \alpha < \log_5 2
\end{cases}
\]
Exercise 32
Compute, as the parameter \(\alpha \in \mathbb{R}\) varies, the limit of the sequence
\[
\lim_{n \rr \ip} (e^{(2-\alpha)\:n}+1)\log{\left(1+\frac{1}{e^n}\right)}
\]
Solution
We have
\[
\begin{split}
a_n=(e^{(2-\alpha)\:n}+1)\log{\left(1+\frac{1}{e^n}\right)} &= (e^{(2-\alpha)\:n}+1)\frac{e^n}{e^n}\log{\left(1+\frac{1}{e^n}\right)} \\
&= (e^{(2-\alpha)\:n}+1)\frac{1}{e^n}\log{\left(1+\frac{1}{e^n}\right)^{e^n}} \\ &\thicksim (e^{(2-\alpha)\:n}+1)\frac{1}{e^n}
\end{split}
\]
since
\[
\left(1+\frac{1}{e^n}\right)^{e^n} \rr e {\rm ~~~~and~~~~} \log{\left(1+\frac{1}{e^n}\right)^{e^n}} \rr 1
\]
Hence
\[
\lim_{n \rr \ip} a_n = \lim_{n \rr \ip} (e^{(2-\alpha)\:n}+1)\frac{1}{e^n}
\]
On the other hand
\[
(e^{(2-\alpha)\:n}+1)\frac{1}{e^n} = e^{(1-\alpha)n}+e^{-n} \rr \begin{cases}
+\infty & \text{if } 1-\alpha>0 \\
1 & \text{if } 1-\alpha=0\\
0 & \text{if } 1-\alpha<0
\end{cases}
\]
Summarizing, we obtain
\[
\lim_{n \rr \ip} (e^{(2-\alpha)\:n}+1)\log{\left(1+\frac{1}{e^n}\right)} = \begin{cases}
+\infty & \text{if } \alpha<1 \\
1 & \text{if } \alpha=1\\
0 & \text{if } \alpha>1
\end{cases}
\]
8. Multiple choice exercises
Exercise 33
Let
\[
\lim_{n\to+\infty}\left(\frac{\sqrt{3}+n}{e-2n}\right)^{n}=\ell
\]
Solution
Setting
\[
a_{n}=\frac{\sqrt{3}+n}{e-2n}
\]
we have \(a_{n}<0\) for \(n\geq2\) and
\[
\left|a_{n}\right|^{n}=\left(\frac{\sqrt{3}+n}{2n-e}\right)^{n}\to0
\]
since \(|a_{n}|\to1/2\) . From \(\left|a_{n}\right|^{n}\to0\) it follows that \((a_{n})^{n}\to0\) . The correct answer is (a).
Exercise 34
Let
\[
\lim_{n\to+\infty}(-1)^{n}n^{\alpha}\left(\frac{1+3n}{2-n}\right)^{n}=\ell
\]
(a) \(\ell\) does not exist if \(\alpha\geq0\)
(b) \(\ell=+\infty\) for every \(\alpha\in\R\)
(c) \(\ell=0\) if \(\alpha<0\)
(d) None of the other answers is correct.
Solution
We set
\[
a_{n}=(-1)^{n}\left(\frac{1+3n}{2-n}\right)^{n}=\left(\frac{1+3n}{n-2}\right)^{n}.
\]
We have \(a_{n}\to+\infty\) since \((1+3n)/(n-2)\to3\) . Let us compare \(a_{n}\) with \(3^{n}\) :
\[
\frac{a_{n}}{3^{n}}=\left(\frac{1+3n}{3n-6}\right)^{n}=\left[\left(1+\frac{1}{\frac{3n-6}{7}}\right)^{\frac{3n-6}{7}}\right]^{\frac{7n}{3n-6}}
\to e^{7/3}
\]
from which
\[
a_{n}\sim3^{n} e^{7/3}.
\]
It follows that
\[
\lim_{n\to+\infty}(-1)^{n}n^{\alpha}\left(\frac{1+3n}{2-n}\right)^{n}=
\lim_{n\to+\infty}e^{7/3}n^{\alpha}3^{n}=+\infty
\]
even when \(\alpha<0\) , because \(3^{n}\) is an infinity of higher order than any power of \(n\) . The correct answer is (b).
Exercise 35
Let
\[
\lim_{n\to+\infty}\left(\frac{an+2}{en+a}\right)^{n}=\ell
\]
(a) \(\ell=e^{ \left(\frac{2}{e}-1\right)}\) if \(a=e\)
(b) \(\ell=1\) if \(a\leq e\)
(c) \(\ell=0\) if \(a>0\)
(d) None of the other answers is correct.
Solution
Setting
\[
a_{n}=\frac{an+2}{en+a}
\]
we have \(a_{n}\to a/e\) , hence
\[
(a_{n})^{n}\to\left\{\begin{array}{lr}+\infty,\ &a>e\\
\\
0,\ &0<a<e\end{array}\right.
\]
therefore answers (b) and (c) are not correct. In the case \(a=e\) we have an indeterminate form \(1^{\infty}\) :
\[
\left(\frac{en+2}{en+e}\right)^{n}=\left[\left(1+\frac{1}{\frac{en+e}{2-e}}\right)^{\frac{en+e}{2-e}}\right]^{\frac{(2-e)n}{en+e}}
\to e^{-1+\frac{2}{e}}.
\]
The correct answer is (a).