91Ó°ÊÓ

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 $x→∞$if, both the functions $f\left(x\right)$and $g\left(x\right)$grow without bound as $x→∞$and also$\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=∞$.

$$\begin{gathered} \underset{x→∞}{\lim }f\left(x\right)=\underset{x→∞}{\lim }A{e}^{kx} \\ =A{e}^{k\left(∞\right)}(\sin ce{e}^{∞}=∞) \\ =∞ \\ \underset{x→∞}{\lim }g\left(x\right)=\underset{x→∞}{\lim }B{x}^r \\ =B{(∞)}^r=∞ \\ sothefunctionf\left(x\right)andg\left(x\right)growwithoutboundx→∞ \end{gathered}$$

$$\begin{gathered} \underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)} \\ \underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x→∞}{\lim }\frac{A{e}^{kx}}{B{x}^r}\left(\mathrm{which}\mathrm{is}\mathrm{in}\frac{∞}{∞}\mathrm{form}\mathrm{as}x→∞\right) \\ theL\text{'}Hospital\text{'}srulestatesthatifthevalueof\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}is\frac{∞}{∞}asx→∞ \\ \underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x→∞}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)} \\ \underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x→∞}{\lim }\frac{Ak{e}^{kx}}{Br{x}^{r-1}}\left(ApplyingL\text{'}Hospital\text{'}sruleof\frac{∞}{∞}form\right) \\ =\underset{x→∞}{\lim }\frac{A{k}^2{e}^{kx}}{Br(r-1)x^{r-2}}\left(ApplyingL\text{'}Hospital\text{'}sruleof\frac{∞}{∞}form\right) \\ =\underset{x→∞}{\lim }\frac{A{k}^3{e}^{kx}}{Br(r-1)(r-2)x^{r-3}}\left(ApplyingL\text{'}Hospital\text{'}sruleof\frac{∞}{∞}form\right) \end{gathered}$$

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

Therefore, exponential growth functions always dominate the power functions.