Q. 76 - 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.

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

Consider the given question,

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

Step 2. Use the limits.

Consider the limits,

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

Using L' Hospitals rule,

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

Using L' Hospitals rule for $100$ times,

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