Question

Use L’Hôpital’s rule to prove that power functions with positive powers always dominate logarithmic functions.

Short answer

Ans: $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{A{x}^r}{\ln Bx}$

Every power functions with positive powers always dominate logarithmic functions.

Step-by-step solution

Step 1. Given Information:

The objective is to prove that every power functions with positive powers always dominate logarithmic functions.

A function $f\left(x\right)$is said to be dominates another function $g\left(x\right)$as $x\to \infty$if, both the functions $f\left(x\right)$and $g\left(x\right)$grow without bound as $x\to \infty$and also$\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$ .

Step 2. Evaluating the values to prove:

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }A{x}^r \\ =A{\left(\infty \right)}^r \\ =\infty \\ \underset{x\to \infty }{\lim }g\left(x\right)=\underset{x\to \infty }{\lim }\left(\ln Bx\right) \\ =\infty \\ so,thefunctionf\left(x\right)andg\left(x\right)growwithoutboundx\to \infty \end{gathered}$$

Step 3. Calculating the values of the limit:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)} \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{A{x}^r}{\ln Bx}\left(\mathrm{whichisin}\frac{\infty }{\infty }\mathrm{formas}x\to \infty \right).....\left(1\right) \\ theL\text{'}Hospital\text{'}srulestatesthatifthevalueof\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}is\frac{\infty }{\infty }asx\to \infty \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)} \end{gathered}$$

Step 4. Applying L' Hospital rule on the right-hand side: 

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{Ar{x}^{r-1}}{{\displaystyle \frac{1}{Bx}}·B}\left(ApplyingL\text{'}Hospital\text{'}sruleof\frac{\infty }{\infty }form\right) \\ =\underset{x\to \infty }{\lim }Ar{x}^{r-1}·x\left(ApplyingL\text{'}Hospital\text{'}sruleof\frac{\infty }{\infty }form\right) \\ =\underset{x\to \infty }{\lim }Ar{x}^{r-1+1}\left(ApplyingL\text{'}Hospital\text{'}sruleof\frac{\infty }{\infty }form\right) \\ =\underset{x\to \infty }{\lim }Ar{x}^r \\ =Ar{(\infty )}^r=\infty \\ \therefore \mathrm{every}\mathrm{power}\mathrm{functions}\mathrm{with}\mathrm{a}\mathrm{positive}\mathrm{powers}\mathrm{always}\mathrm{dominate}\mathrm{logarithmic}\mathrm{functions}. \end{gathered}$$