A fundamental question that arises regarding infinite sequences is the behavior of the terms as \(n\) gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as \(n \rightarrow \infty\). For example, consider the following four sequences and their different behaviors as \(n \rightarrow \infty\).
Definition
Given a sequence \(\left\{a_n\right\}\), if the terms \(a_n\) become arbitrarily close to a finite number \(L\) as \(n\) becomes sufficiently large, we say \(\left\{a_n\right\}\) is convergent sequence and \(L\) is the limit of the sequence. In this case, we write
\(
\lim _{n \rightarrow \infty} a_n=L
\)
If a sequence \(\left\{a_n\right\}\) is not convergent, we say it is a divergent sequence.
Theorem: Limit of a Sequence defined by a Function
Consider a sequence \(\left\{a_n\right\}\) such that \(a_n=f(n)\) for all \(n \geq 1\). If there exists a real number \(L\) such that
\(
\lim _{x \rightarrow \infty} f(x)=L,
\)
then \(\left\{a_n\right\}\) converges and
\(
\lim _{n \rightarrow \infty} a_n=L .
\)
Theorem: Algebraic Limit Laws
Given sequences \(\left\{a_n\right\}\) and \(\left\{b_n\right\}\) and any real number \(c\), if there exist constants \(A\) and \(B\) such that \(\lim _{n \rightarrow \infty} a_n=A\) and \(\lim _{n \rightarrow \infty} b_n=B\), then
End Behavior of Rational Functions
The value of \(\lim _{x \rightarrow \infty} \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials can be determined by looking at the degrees of the numerator and denominator.
Example 1: For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.
(i) \(\left\{5-\frac{3}{n^2}\right\}\)
(ii) \(\left\{\frac{3 n^4-7 n^2+5}{6-4 n^4}\right\}\)
(iii) \(\left\{\frac{2^n}{n^2}\right\}\)
(iv) \(\left\{\left(1+\frac{4}{n}\right)^n\right\}\)
Solution: (i) We know that \(\frac{1}{n} \rightarrow 0\). Using this fact, we conclude that
\(
\lim _{n \rightarrow \infty} \frac{1}{n^2}=\lim _{n \rightarrow \infty}\left(\frac{1}{n}\right) \cdot \lim _{n \rightarrow \infty}\left(\frac{1}{n}\right)=0 .
\)
Therefore,
\(
\lim _{n \rightarrow \infty}\left(5-\frac{3}{n^2}\right)=\lim _{n \rightarrow \infty} 5-3 \lim _{n \rightarrow \infty} \frac{1}{n^2}=5-3.0=5 .
\)
The sequence converges and its limit is 5 .
(ii) By factoring \(n^4\) out of the numerator and denominator and using the limit laws above, we have
\(
\begin{aligned}
\lim _{n \rightarrow \infty} \frac{3 n^4-7 n^2+5}{6-4 n^4} & =\lim _{n \rightarrow \infty} \frac{3-\frac{7}{n^2}+\frac{5}{n^4}}{\frac{6}{n^4}-4} \\
& =\frac{\lim _{n \rightarrow \infty}\left(3-\frac{7}{n^2}+\frac{5}{n^4}\right)}{\lim _{n \rightarrow \infty}\left(\frac{6}{n^4}-4\right)}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{\left(\lim _{n \rightarrow \infty}(3)-\lim _{n \rightarrow \infty} \frac{7}{n^2}+\lim _{n \rightarrow \infty} \frac{5}{n^4}\right)}{\left(\lim _{n \rightarrow \infty} \frac{6}{n^4}-\lim _{n \rightarrow \infty}(4)\right)} \\
& =\frac{\left(\lim _{n \rightarrow \infty}(3)-7 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^2}+5 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^4}\right)}{\left(6 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^4}-\lim _{n \rightarrow \infty}(4)\right)} \\
& =\frac{3-7 \cdot 0+5 \cdot 0}{6 \cdot 0-4}=-\frac{3}{4} .
\end{aligned}
\)
The sequence converges and its limit is \(\frac{-3}{4}\).
(iii) Consider the related function \(f(x)=\frac{2^x}{x^2}\) defined on all real numbers \(x>0\). Since \(2^x \rightarrow \infty\) and \(x^2 \rightarrow \infty\) as \(x \rightarrow \infty\), apply L’Hôpital’s rule and write
\(
\begin{array}{rlrl}
\lim _{x \rightarrow \infty} \frac{2^x}{x^2} & =\lim _{x \rightarrow \infty} \frac{2^x \ln 2}{2 x} & & \text { Take the derivatives of the } n \\
& =\lim _{x \rightarrow \infty} \frac{2^x(\ln 2)^2}{2} & & \text { Take the derivatives again. } \\
& =\infty . &
\end{array}
\)
We conclude that the sequence diverges.
(iv) Consider the function \(f(x)=\left(1+\frac{4}{x}\right)^x\) defined on all real numbers \(x>0\). This function has the indeterminate form \(1^{\infty}\) as \(x \rightarrow \infty\). Let
\(
y=\lim _{x \rightarrow \infty}\left(1+\frac{4}{x}\right)^x
\)
Now taking the natural logarithm of both sides of the equation, we obtain
\(
\ln (y)=\ln \left[\lim _{x \rightarrow \infty}\left(1+\frac{4}{x}\right)^x\right] .
\)
Since the function \(f(x)=\ln x\) is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,
\(
\ln (y)=\lim _{x \rightarrow \infty}\left[\ln \left(1+\frac{4}{x}\right)^x\right] .
\)
Using properties of logarithms, we write
\(
\lim _{x \rightarrow \infty}\left[\ln \left(1+\frac{4}{x}\right)^x\right]=\lim _{x \rightarrow \infty} x \ln \left(1+\frac{4}{x}\right) .
\)
Since the right-hand side of this equation has the indeterminate form \(\infty \cdot 0\), rewrite it as a fraction to apply L’Hôpital’s rule. Write
\(
\lim _{x \rightarrow \infty} x \ln \left(1+\frac{4}{x}\right)=\lim _{x \rightarrow \infty} \frac{\ln \left(1+\frac{4}{x}\right)}{\frac{1}{x}} .
\)
Since the right-hand side is now in the indeterminate form \(\frac{0}{0}\), we are able to apply L’Hôpital’s rule. We conclude that
\(
\lim _{x \rightarrow \infty} \frac{\ln \left(1+\frac{4}{x}\right)}{\frac{1}{x}}=\lim _{x \rightarrow \infty} \frac{4}{1+\frac{4}{x}}=4 .
\)
Therefore, \(\ln (y)=4\) and \(y=e^4\). Therefore, since \(\lim _{x \rightarrow \infty}\left(1+\frac{4}{x}\right)^x=e^4\), we can conclude that the sequence \(\left\{\left(1+\frac{4}{n}\right)^n\right\}\) converges to \(e^4\).
L’HÔPITAL’S RULE
Suppose \(f\) and \(g\) are differentiable functions over an open interval \((a, \infty)\) for some value of \(a\). If either:
Limit Using Series Expansion:
Expansion of function like binomial expansion, exponential & logarithmic expansion, expansion of \(\sin x, \cos x\), tan \(x\) should be remembered by heart which are given below :
Example 2: \(\lim _{x \rightarrow 0} \frac{e^x-e^{-x}-2 x}{x-\sin x}\)
Solution:
[/latex]
\lim _{x \rightarrow 0} \frac{e^x-e^{-x}-2 x}{x-\sin x} \Rightarrow \lim _{x \rightarrow 0} \frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \ldots-\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots . .\right)-2 x}{x-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!} \ldots .\right)}
[/latex]
\(
\Rightarrow \lim _{x \rightarrow 0} \frac{2 \cdot \frac{x^3}{6}+2 \cdot \frac{x^5}{5!}+\ldots \ldots}{\frac{x^3}{6}+\frac{x^5}{5!} \ldots .}
\)
\(
\Rightarrow \quad \lim _{x \rightarrow 0} \frac{x^3\left(\frac{1}{3}+\frac{1}{60} x^2+\ldots .\right)}{x^3\left(\frac{1}{6}+\frac{1}{120} x^2+\ldots . .\right)}=\frac{1 / 3}{1 / 6}=2
\)
Example 3: Evaluate : \(\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}\)
Solution:
\(
\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}=\lim _{x \rightarrow 0} \frac{e^x \times e^{(\tan x-x)}-e^x}{\tan x-x}
\)
\(
=\lim _{x \rightarrow 0} \frac{e^x\left(e^{\tan x-x}-1\right)}{\tan x-x}=\lim _{y \rightarrow 0} \frac{e^x\left(e^y-1\right)}{y} \text { where } y=\tan x-x \text { and } \lim _{y \rightarrow 0} \frac{e^y-1}{y}=1
\)
\(
=e^0 \times 1 \quad \text { [as } x \rightarrow 0, \tan x-x \rightarrow 0 \text { ] }
\)
\(
=1 \times 1=1
\)
Note1: \(\operatorname{Lim}_{x \rightarrow 0} \frac{a^x-1}{x}=\ln a(a>0) \text { In particular } \operatorname{Lim}_{x \rightarrow 0} \frac{e^x-1}{x}=1 .\)
\(
\text { In general if } \operatorname{Lim}_{x \rightarrow a} f(x)=0 \text {, then } \operatorname{Lim}_{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)}=\ell \text { na, } a>0
\)
Note2: \(\operatorname{Lim}_{x \rightarrow 0}(1+x)^{1 / x}=e=\operatorname{Lim}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x\) (Note : The base and exponent depends on the same variable.)
In general, if \(\operatorname{Lim}_{x \rightarrow a} f(x)=0\), then \(\operatorname{Lim}_{x \rightarrow a}(1+f(x))^{1 / f(x)}=e\)
\(
\operatorname{Lim}_{x \rightarrow 0} \frac{\ln (1+x)}{x}=1
\)
\(
\text { If } \operatorname{Lim}_{x \rightarrow a} f(x)=1 \text { and } \operatorname{Lim}_{x \rightarrow a} \phi(x)=\infty \text {, then } ; \operatorname{Lim}_{x \rightarrow a}[f(x)]^{\phi(x)}=e^k \text { where } k=\operatorname{Lim}_{x \rightarrow a} \phi(x)[f(x)-1]
\)
Example 4: Evaluate \(\operatorname{Lim}_{x \rightarrow 1}\left(\log _3 3 x\right)^{\log _x 3}\)
Solution:
\(
\operatorname{Lim}_{x \rightarrow 1}\left(\log _3 3 x\right)^{\log _x 3}=\operatorname{Lim}_{x \rightarrow 1}\left(\log _3 3+\log _3 x\right)^{\log _x 3}
\)
\(
=\operatorname{Lim}_{x \rightarrow 1}\left(1+\log _3 x\right)^{1 / \log _3 x}=e \quad \because \quad \log _{ b } a =\frac{1}{\log _{ a } b }
\)
Example 5: Evaluate: \(\operatorname{Lim}_{x \rightarrow 0} \frac{x \ln (1+2 \tan x)}{1-\cos x}\)
Solution:
\(\operatorname{Lim}_{x \rightarrow 0} \frac{x \ln (1+2 \tan x)}{1-\cos x}=\operatorname{Lim}_{x \rightarrow 0} \frac{x \ln (1+2 \tan x)}{\frac{1-\cos x}{x^2} \cdot x^2} \cdot \frac{2 \tan x}{2 \tan x}=4\)
Example 6: Evaluate \(: \lim _{x \rightarrow \infty}\left(\frac{2 x^2-1}{2 x^2+3}\right)^{4 x^2+2}\)
Solution: Since it is in the form of \(1^{\circ}\)
\(
\lim _{x \rightarrow \infty}\left(\frac{2 x^2-1}{2 x^2+3}\right)^{4 x^2+2}
\)
\(
=e^{\lim _{x \rightarrow \infty}} \left(\frac{2 x^2-1-2 x^2-3}{2 x^2+3}\right) \left(4 x^2+2\right)=e^{-8}
\)
Example 7: \(\lim _{x \rightarrow 0}\left(\frac{a^x+b^x+c^x}{3}\right)^{1 / x}=?\)
Solution:
\(
\lim _{x \rightarrow 0}\left(\frac{a^x+b^x+c^x}{3}\right)^{1 / x}=\lim _{x \rightarrow 0}\left(1+\frac{a^x+b^x+c^x-3}{3}\right)^{1 / x}
\)
\(
=\lim _{x \rightarrow 0}\left[\left(1+\frac{\left(a^x-1\right)}{3}+\frac{\left(b^x-1\right)}{3}+\frac{\left(c^x-1\right)}{3}\right)^{\frac{3}{\left(a^x-1\right)+\left(b^x-1\right)+\left(c^x-1\right)}}\right]^{\frac{a^x-1+b^x-1+c^x-1}{3 x}}
\)
\(
=e^{1 / 3} \lim _{x \rightarrow 0}\left[\frac{a^x-1}{x}+\frac{b^x-1}{x}+\frac{c^x-1}{x}\right]=e^{{1 / 3}(\log a+\log b+\log c)}=e^{\log (a b c) 1 / 3}=( abc )^{1 / 3}
\)
Example 8: Evaluate : \(\lim _{x \rightarrow \infty}\left(\frac{7 x^2+1}{5 x^2-1}\right)^{\frac{x^5}{1-x^3}}\)
Solution: Here \(f ( x )=\frac{7 x ^2+1}{5 x ^2-1}, \quad \phi( x )=\frac{ x ^5}{1- x ^3}=\frac{ x ^2 \cdot x ^3}{1- x ^3}=\frac{ x ^2}{\frac{1}{ x ^3}-1}\)
\(
\therefore \quad \lim _{x \rightarrow \infty} f(x)=\frac{7}{5} \quad \& \quad \lim _{x \rightarrow \infty} \phi(x) \rightarrow-\infty
\)
\(
\Rightarrow \quad \lim _{x \rightarrow \infty}(f(x))^{\phi(x)}=\left(\frac{7}{5}\right)^{-\infty}=0
\)
Note: \(\text { If } \operatorname{Lim}_{x \rightarrow a} f(x)=A>0 \& \operatorname{Lim}_{x \rightarrow a} \phi(x)=B \text { (a finite quantity) then } ; \operatorname{Lim}_{x \rightarrow a}[f(x)]^{\phi(x)}=e^{B \ln A}=A^B\)
Example 9: Evaluate : \(\lim _{x \rightarrow 0} \frac{\cos ^2\left\{1-\cos ^2\left(1-\cos ^2\left(\ldots \ldots \ldots . .\left(1-\cos ^2(x)\right)\right)\right)\right\}}{\sin \left[\pi\left(\frac{\sqrt{x+4}-2}{x}\right)\right]}\)
Solution: Let \(A=\lim _{x \rightarrow 0} \frac{\cos ^2\left\{1-\cos ^2\left(1-\cos ^2\left(\ldots \ldots \ldots \ldots\left(1-\cos ^2(x)\right)\right)\right)\right\}}{\sin \left[\pi\left(\frac{\sqrt{x+4}-2}{x}\right)\right]}\)
\(
=\lim _{x \rightarrow 0} \frac{\cos ^2\left\{\sin ^2\left(\sin ^2\left(\ldots \ldots .\left(1-\cos ^2(x)\right)\right)\right)\right\}}{\sin \left(\pi\left(\frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\right)\right)}=\frac{\cos ^2 0}{1} \lim _{x \rightarrow 0} \frac{1}{\sin \left(\pi \frac{(x+4-4)}{x(\sqrt{x+4}+2)}\right)}=\frac{1}{\sin \frac{\pi}{4}}=\sqrt{2}
\)
Example 10: Evaluate the following limits, if exist \(\lim _{n \rightarrow \infty} n^{-n^2}\left((n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots \ldots \ldots\left(n+\frac{1}{2^{n-1}}\right)\right)^n\)
Solution:
\(
\lim _{n \rightarrow \infty} n^{-n^2}\left((n+1)\left(n+\frac{1}{2}\right) \ldots \ldots\left(n+\frac{1}{2^{n-1}}\right)\right)^n=\lim _{n \rightarrow \infty}\left(\frac{(n+1)\left(n+\frac{1}{2}\right) \ldots \ldots\left(n+\frac{1}{2^{n-1}}\right)}{n^n}\right)^n
\)
\(
=\lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right)^n \cdot\left(\frac{n+\frac{1}{2}}{n}\right)^n \cdots \cdots \cdot\left(\frac{n+\frac{1}{2^{n-1}}}{n}\right)^n
\)
\(
=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n \cdot\left(1+\frac{1}{2 n}\right)^{\frac{2 n}{2}} \cdots \cdots \cdot\left(1+\frac{1}{2^{n-1} n}\right)^{\frac{2^{n-1} n}{2^{n-1}}}
\)
\(
=e \cdot e^{\frac{1}{2}} \cdot e^{\frac{1}{4}} \ldots \ldots=e^{\left(1+\frac{1}{2}+\frac{1}{4} \cdots \cdots\right)}=e^2
\)
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