Q. 93 - Calculus

Question

Use L’Hoˆpital’s rule to prove that exponential growth functions always dominate power functions.

Step-by-Step Solution

Answer

Ans: Exponential growth functions always dominate the power functions. 

Step 1. Given Information:

The objective is to prove that every exponential growth function dominates the power 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:

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }A{e}^{kx} \\ =A{e}^{k\left(\infty \right)} (\sin ce e^{\infty }=\infty ) \\ =\infty \\ \underset{x\to \infty }{\lim }g\left(x\right)=\underset{x\to \infty }{\lim }B{x}^r \\ =B{(\infty )}^r =\infty \\ so the function f\left(x\right) and g\left(x\right) grow without bound x\to \infty \end{gathered}$$

Step 3. Calculating the values of the limit:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)} \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{A{e}^{kx}}{B{x}^r} \left( \mathrm{which}\mathrm{is}\mathrm{in} \frac{\infty }{\infty } \mathrm{form}\mathrm{as} x\to \infty \right) \\ the L\text{'}Hospital\text{'}s rule states that if the value of\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)} is \frac{\infty }{\infty } as x\to \infty \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)} \\ \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{Ak{e}^{kx}}{Br{x}^{r-1}} \left(Applying L\text{'}Hospital\text{'}s rule of \frac{\infty }{\infty } form\right) \\ =\underset{x\to \infty }{\lim }\frac{A{k}^2{e}^{kx}}{Br(r-1)x^{r-2}} \left(Applying L\text{'}Hospital\text{'}s rule of \frac{\infty }{\infty } form\right) \\ =\underset{x\to \infty }{\lim }\frac{A{k}^3{e}^{kx}}{Br(r-1)(r-2)x^{r-3}} \left(Applying L\text{'}Hospital\text{'}s rule of \frac{\infty }{\infty } form\right) \end{gathered}$$

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

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{A{k}^r{e}^{kx}}{Br(r-1)(r-2)(r-3)\cdots \left(2\right)\left(1\right)x^{r-r}} \\ =\underset{x\to \infty }{\lim }\frac{A{k}^r{e}^{kx}}{Br(r-1)(r-2)(r-3)\cdots \left(2\right)\left(1\right)}\left( since x^0=1\right) \\ =\frac{\infty }{C}\left(\begin{array}{l} Take \mathrm{the} \mathrm{value} Br(r-1)(r-2)(r-3)\cdots \left(2\right)\left(1\right) as \mathrm{some} \\ constant \mathrm{say} C\end{array}\right) \\ =\infty \end{gathered}$$

Therefore, exponential growth functions always dominate the power functions.