Question

(Requires calculus) For each of these pairs of functions, determine whether f and g are asymptotic

1. $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{,}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{x}$
   
2. $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{2}}^{\mathbf{x}\mathbf{+}\mathbf{3}}\mathbf{,}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{2}}^{\mathbf{x}\mathbf{+}\mathbf{7}}$
   
3. $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{2}}^{{\mathbf{2}}^{\mathbf{x}}}\mathbf{,}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{2}}^{{\mathbf{x}}^{\mathbf{2}}}$
   
4. $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{2}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}}\mathbf{,}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{2}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}}$
   

Short answer

If$\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=1$, then f and g are said to be asymptotic.

If $\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}\ne 1$including limit does not exist, then f and g are said to be not asymptotic.

Step-by-step solution

Subpart (a): f(x)=log⁡(x2+1),g(x)=log⁡xStep 1: Calculating the Asymptotic condition:

Given,

$\begin{array}{r}f\left(x\right)=\log \left(x^2+1\right)\\ g\left(x\right)=\log x\end{array}$

Then,

$\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{\log \left(x^2+1\right)}{\log x}\\ =\underset{x\to \mathrm{\infty }}{lim} \frac{\log \left(1+\frac{1}{x^2}\right)+\log \left(x^2\right)}{\log x}\\ =\underset{x\to \mathrm{\infty }}{lim} \left[\frac{\log \left(1+\frac{1}{x^2}\right)}{\log x}+\frac{\log \left(x^2\right)}{\log x}\right]\\ =\underset{x\to \mathrm{\infty }}{lim} \frac{\log \left(1+\frac{1}{x^2}\right)}{\log x}+\underset{x\to \mathrm{\infty }}{lim} \frac{2\log x}{\log x}\end{array}$

Applying limit x→∞:

Since$\frac{1}{\mathrm{\infty }}=0$,

$\begin{array}{r}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=0+2\\ =2\end{array}$

Therefore, $\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}\ne 1\\ \end{array}$so, f and g are not asymptotic.

Subpart (b): f(x)=2x+3,g(x)=2x+7Step 1: Calculating the Asymptotic condition:

Given,

$\begin{array}{r}f\left(x\right)=2^{x+3}\\ g\left(x\right)=2^{x+7}\end{array}$

Then,

$\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{2^{x+3}}{2^{x+7}}\\ \because \left(\frac{a^m}{a^n}=a^{m-n}\right)\\ =\underset{x\to \mathrm{\infty }}{lim} \left(2^{x+3-x-7}\right)\\ =\underset{x\to \mathrm{\infty }}{lim} 2^{-4}\end{array}$

Applying limit x→∞:

Since$\frac{1}{\mathrm{\infty }}=0$,

$\begin{array}{r}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\frac{1}{2^4}\\ =\frac{1}{16}\end{array}$

Thus, $\begin{array}{r}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}\ne 1\\ \end{array}$so, f and g are not asymptotic.

Subpart (c): f(x)=22x,g(x)=2x2Step 1: Calculating the Asymptotic condition:

Given,

$f\left(x\right)=2^{2^x},g\left(x\right)=2^{x^2}$

Then,

$\begin{array}{c}\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{2^{2^x}}{2^{x^2}}\\ \left(\because \frac{a^m}{a^n}=a^{m-n}\right)\\ =\underset{x\to \mathrm{\infty }}{lim} 2^{\left(2^x-x^2\right)}\end{array}$

Applying limit x→∞:

Since$\frac{1}{\mathrm{\infty }}=0$ and$a^{\infty }=\infty$

data-custom-editor="chemistry" $\begin{array}{r}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=2^{\mathrm{\infty }}\\ =\mathrm{\infty }\end{array}$

Thus, data-custom-editor="chemistry" $\begin{array}{r}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}\ne 1\\ \end{array}$so, f and g are not asymptotic.

Subpart (d): f(x)=2x2+x+1,g(x)=2x2+2xStep 1: Calculating the Asymptotic condition:

Given,

$\begin{array}{r}f\left(x\right)=2^{x^2+x+1}\\ g\left(x\right)=2^{x^2+2x}\end{array}$

Then,

$\begin{array}{c}\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{2^{x^2+x+1}}{2^{x^2+2x}}\\ \left(\because \frac{a^m}{a^n}=a^{m-n}\right)\\ =\underset{x\to \mathrm{\infty }}{lim} 2^{x^2+x+1-x^2-2x}\\ =\underset{x\to \mathrm{\infty }}{lim} 2^{1-x}\\ \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} 2^{\frac{1}{x}\left(x-x^2\right)}\end{array}$

Applying limit x→∞:

Since $\frac{1}{\mathrm{\infty }}=0$,

$\begin{array}{c}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=2^0\\ =1\end{array}$

Thus,$\begin{array}{c}\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}\ne 1\\ \end{array}$so, f and g are asymptotic.