Q. 77 - Calculus

Question

As you will prove in Exercises 93 and 94, exponential growth functions ${\mathrm{e}}^{\mathrm{kx}}$always dominate power functions $x^r$, and power functions $x^r$with positive powers always dominate logarithmic functions ${\log }_{b}x$. Use these facts to quickly determine each of the limits in Exercises 75-80.

$\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}$

Step-by-Step Solution

Answer

The limit is $\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\infty$.

Step 1. Given information.

Consider the given question,

$\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}$

Step 2. Use the limits.

Consider the limits,

$\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{1}{2\sqrt{x}}}}{{\displaystyle \frac{300}{x}}}$

Using L' Hospitals rule,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\underset{x\to \infty }{\lim }\left(\frac{1}{2\sqrt{x}}·\frac{x}{300}\right) \\ \underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\underset{x\to \infty }{\lim }\left(\frac{\sqrt{x}}{600}\right) \\ \underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\infty \end{gathered}$$