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

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$\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$ .

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

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

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