Q. 94 - Calculus

Question

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

Step-by-Step Solution

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 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:

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }A{x}^r \\ =A{\left(\infty \right)}^r \\ =\infty \\ \underset{x\to \infty }{\lim }g\left(x\right)=\underset{x\to \infty }{\lim }\left(\ln Bx\right) \\ =\infty \\ so, the function f\left(x\right) and g\left(x\right) grow without bound x\to \infty \end{gathered}$$

Step 3. Calculating the values of the limit:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)} \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{A{x}^r}{\ln Bx} \left( \mathrm{whichisin} \frac{\infty }{\infty } \mathrm{formas} x\to \infty \right) .....\left(1\right) \\ the L\text{'}Hospital\text{'}s rule states that if the value of\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)} is \frac{\infty }{\infty } as x\to \infty \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)} \end{gathered}$$

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

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