Chapter 4.6, Problem 3E

a.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\infty$

Explanation of Solution

Given information:

  $x_n=10n$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=10n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(10n\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\infty \end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ does not exist.

b.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$

Explanation of Solution

Given information:

  $x_n=\frac{10}{n}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{10}{n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{10}{n}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\left(\frac{10}{n}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$ .

c.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{4}{5}$

Explanation of Solution

Given information:

  $x_n=\frac{4n^2+7n+12}{11-n+5n^2}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{4n^2+7n+12}{11-n+5n^2}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{4n^2+7n+12}{11-n+5n^2}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\left(\frac{4\frac{n^2}{n^2}+7\frac{n}{n^2}+12\frac{1}{n^2}}{11\frac{1}{n^2}-\frac{n}{n^2}+5\frac{n^2}{n^2}}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\frac{4+7\left(\frac{1}{n}\right)+12\frac{1}{n^2}}{11\frac{1}{n^2}-\frac{1}{n}+5}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{4}{5}\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{4}{5}$ .

d.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\infty$

Explanation of Solution

Given information:

  $x_n=\frac{6n^3+5n^2+3n+8}{10n^2+7n+5}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{6n^3+5n^2+3n+8}{10n^2+7n+5}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{6n^3+5n^2+3n+8}{10n^2+7n+5}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\frac{6+5\left(\frac{1}{n}\right)+3\left(\frac{1}{n^2}\right)+8\left(\frac{1}{n^3}\right)}{10\left(\frac{1}{n}\right)+7\left(\frac{1}{n^2}\right)+5\left(\frac{1}{n^3}\right)}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{6}{0}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\infty \end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ does not exist .

e.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$

Explanation of Solution

Given information:

  $x_n=\frac{8n^2+4n+1}{11n^3-n+5}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{8n^2+4n+1}{11n^3-n+5}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{8n^2+4n+1}{11n^3-n+5}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\frac{8\left(\frac{1}{n}\right)+4\left(\frac{1}{n^2}\right)+1\left(\frac{1}{n^3}\right)}{11-\left(\frac{1}{n^2}\right)+5\left(\frac{1}{n^3}\right)}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{0}{11}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$ .

f.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{3}{5}$

Explanation of Solution

Given information:

  $x_n=\frac{6n^3+5n^2+3n+8}{10n^3+7n^2+5n-8}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{6n^3+5n^2+3n+8}{10n^3+7n^2+5n-8}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{6n^3+5n^2+3n+8}{10n^3+7n^2+5n-8}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\frac{6+5\left(\frac{1}{n}\right)+3\left(\frac{1}{n^2}\right)+8\left(\frac{1}{n^3}\right)}{10+7\left(\frac{1}{n}\right)+5\left(\frac{1}{n^2}\right)-8\left(\frac{1}{n^3}\right)}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{6}{10}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{3}{5}\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{3}{5}$ .

g.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=e^2$

Explanation of Solution

Given information:

  $x_n={\left(1+\frac{1}{n}\right)}^{2n}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n={\left(1+\frac{1}{n}\right)}^{2n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}{\left(1+\frac{1}{n}\right)}^{2n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to 0}{{\displaystyle \lim }}{\left(\left(1+\frac{1}{n}\right)\right)}^2\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=e^2\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=e^2$ .

h.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{1}{e}$

Explanation of Solution

Given information:

  $x_n={\left(1+\frac{1}{n}\right)}^{-n}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n={\left(1+\frac{1}{n}\right)}^{-n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}{\left(1+\frac{1}{n}\right)}^{-n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\frac{1}{{\left(1+\frac{1}{n}\right)}^n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{1}{e}\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\frac{1}{e}$ .

i.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$

Explanation of Solution

Given information:

  $x_n={\left(-0.9\right)}^n$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n={\left(-0.9\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}{\left(-0.9\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}{\left(-1\right)}^n{\left(\frac{9}{10}\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0\left(as{9}^n<<{10}^n\right)\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$ .

j.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\infty$

Explanation of Solution

Given information:

  $x_n={\left(-1.1\right)}^n$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n={\left(-1.1\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}{\left(-1.1\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}{\left(-1\right)}^n{\left(\frac{11}{10}\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\infty \left(as{11}^n>>{10}^n\right)\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ does not exist .

k.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\left({\left(-1\right)}^n+1\right)e$

Explanation of Solution

Given information:

  $x_n=\left({\left(-1\right)}^n+1\right){\left(1+\frac{1}{n}\right)}^n$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\left({\left(-1\right)}^n+1\right){\left(1+\frac{1}{n}\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left({\left(-1\right)}^n+1\right){\left(1+\frac{1}{n}\right)}^n\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\left({\left(-1\right)}^n+1\right)e\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\left({\left(-1\right)}^n+1\right)e$ .

l.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$

Explanation of Solution

Given information:

  $x_n=\frac{2^n+3^n}{5^n}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{2^n+3^n}{5^n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\frac{2^n+3^n}{5^n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{2^n}{5^n}+\frac{3^n}{5^n}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0+0\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$ .

m.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$

Explanation of Solution

Given information:

  $x_n=\frac{n!}{n^n}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n=\frac{n!}{n^n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{n\to \infty }{{\displaystyle \lim }}\frac{n!}{n^n}\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}{\left(\frac{1}{n}\right)}^n\left(n!\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0$ .

n.

To determine

Estimate $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ or determine that it does not exist.

Answer to Problem 3E

  $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=1$

Explanation of Solution

Given information:

  $x_n={\left(n\right)}^{\frac{1}{n}}$

Calculation:

Here we will consider:

  $\begin{array}{l}{x}_n={\left(n\right)}^{\frac{1}{n}}\\ \ln x_n=\ln {\left(n\right)}^{\frac{1}{n}}\\ \ln x_n=\left(\frac{1}{n}\right)\ln \left(n\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}\ln x_n=\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{1}{n}\right)\ln \left(n\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=\underset{\frac{1}{n}\to 0}{{\displaystyle \lim }}\left(\frac{1}{n}\right)\ln \left(n\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=0\end{array}$

Now, taking antilog:

  $\begin{array}{l}\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=e^0\\ \underset{n\to \infty }{{\displaystyle \lim }}{x}_n=1\end{array}$

Hence, $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n$ has value $\underset{n\to \infty }{{\displaystyle \lim }}{x}_n=1$ .