Q. 75 - Calculus

Question

As you will prove in Exercises 93 and 94, exponential growth functions $e^{kx}$ 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 $\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=0$.

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,

$\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\underset{x\to \infty }{\lim }\frac{101x^{100}}{e^x}$

Using L' Hospitals rule,

$\underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\underset{x\to \infty }{\lim }\frac{101\times 100\times 99\times ....\times 1}{e^x}$

Using L' Hospitals rule for $100$ times,

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=\frac{1}{\infty } \\ \underset{x\to \infty }{\lim }\frac{x^{101}+500}{e^x}=0 \end{gathered}$$