Examples

L’Hopital’s Rule for sequence limits

(a) What is the limit of $a_n=\frac{\ln n}{n}$?
(b) What is the limit of $b_n=\frac{(\ln n{)}^2}{n}$?
(c) What is the limit of $c_n=n\left(\sqrt{n^2+1}-\sqrt{n^2}\right)$?
(d) What is the limit of $d_n=n{e}^{-n}$?

Solution

(a)

Identify indeterminate form $\frac{\infty }{\infty }$. Change from $n$ to $x$ and apply L’Hopital:

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x}\stackrel{\frac{d}{dx}}{\gg \gg }\underset{x\to \infty }{\lim }\frac{1/x}{1}=0$$

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(b)

Identify indeterminate form $\frac{\infty }{\infty }$. Change from $n$ to $x$ and apply L’Hopital:

$$\underset{x\to \infty }{\lim }\frac{(\ln x{)}^2}{x}\stackrel{\frac{d}{dx}}{\gg \gg }\underset{x\to \infty }{\lim }\frac{2\ln x\cdot \frac{1}{x}}{1}=2\underset{x\to \infty }{\lim }\frac{\ln x}{x}\stackrel{\text{(by (a) result)}}{=}0$$

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(c)

(1) Identify form $\infty \cdot 0$ and rewrite as $\frac{\infty }{\infty }$:

$$n\left(\sqrt{n^2+1}-\sqrt{n^2}\right)\gg \gg \frac{\sqrt{n^2+1}-\sqrt{n^2}}{1/n}$$

(2) Change from $n$ to $x$ and apply L’Hopital:

$$\frac{\sqrt{x^2+1}-\sqrt{x^2}}{1/x}\gg \gg \frac{\frac{1}{2}{\left(x^2+1\right)}^{-1/2}(2x)-1}{-1/x^2}$$

(3) Simplify:

$$\gg \gg \frac{-2x^3}{\sqrt{x^2+1}}+\frac{1}{x^2}\gg \gg \frac{-2x^5+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}$$

(4) Consider the limit:

$$\frac{-2x^5+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}\stackrel{x\to \infty }{⟶}\frac{-2x^5+x}{x^3}⟶-\infty$$

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(d)

(1) Identify form $\infty \cdot 0$ and rewrite as $\frac{\infty }{\infty }$:

$$n{e}^{-n}\gg \gg \frac{n}{e^n}$$

(2) Change $n$ to $x$ and apply L’Hopital:

$$\frac{x}{e^x}\gg \gg \frac{1}{e^n}\stackrel{x\to \infty }{⟶}0$$

Squeeze theorem

Use the squeeze theorem to show that $\frac{4^n}{n!}\to 0$ as $n\to \infty$.

Solution

(1) We will squeeze the given general term above $0$ and below a sequence $b_n$ that we must devise:

$$0\le \frac{4^n}{n!}\le b_n$$

(2) We need $b_n$ to satisfy $b_n\to 0$ and $\frac{4^n}{n!}\le b_n$. Let us study $\frac{4^n}{n!}$.

$$\frac{4^n}{n!}=\frac{4\cdot 4\cdot \cdots \cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4}{n(n-1)\cdots 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$$

(3) Now for the trick. Collect factors in the middle bunch:

$$\frac{4^n}{n!}=\frac{4}{n}\left(\frac{4}{n-1}\cdot \frac{4}{n-2}\cdot \cdots \cdot \frac{4}{7}\cdot \frac{4}{6}\cdot \frac{4}{5}\right)\frac{4\cdot 4\cdot 4\cdot 4}{4\cdot 3\cdot 2\cdot 1}$$

(4) Each factor in the middle bunch is $<1$ so the entire middle bunch is $<1$. Therefore:

$$\frac{4^n}{n!}<\frac{4}{n}\cdot \frac{4^4}{4!}=\frac{1024}{24n}$$

Now we can easily see that $1024/24n\to 0$ as $n\to \infty$, so we set $b_n=1024/24n$ and we are done.

Monotonicity

Show that $a_n=\sqrt{n+1}-\sqrt{n}$ converges.

Solution

(1) Observe that $a_n>0$ for all $n$.

Because $n+1>n$, we know $\sqrt{n+1}>\sqrt{n}$.

Therefore $\sqrt{n+1}-\sqrt{n}>0$

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(2) Show $a_x$ is decreasing:

Replace $n$ with $x$: $a_x=\sqrt{x+1}-\sqrt{x}$ considered as a differentiable function.

Take derivative to show decreasing:

$$\begin{array}{c}\frac{d}{dx}{a}_x=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}\\ \\ \gg \gg \frac{2\left(\sqrt{x}-\sqrt{x+1}\right)}{4\sqrt{x}\sqrt{x+1}}\end{array}$$

Denominator is $>0$. Numerator is $<0$. So $\frac{d}{dx}{a}_x<0$ and $a_x$ is monotone decreasing.

Therefore $a_n$ is monotone decreasing as $n\to \infty$.

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(3) Since $a_n>0$ it is bounded below by $0$. It is monotone decreasing. Conclude that it converges.