Q. 17 - Calculus

Question

To prove that the limit forms in Theorem 1.33 are indeterminate, we need only list explicit examples of limits that do and do not exist for each form. Do so for each of the limit forms

from Exercises 15–21. For the last three forms you may want to experiment with a graphing utility to find your examples.

$\frac{\infty }{\infty }$that approaches (a) 1 (b) 6 (c)

Step-by-Step Solution

Answer

The correct answer is $\left(c\right)\infty$

Step 1. Given Information

The given limit is $\frac{\infty }{\infty }$$0·\infty$

Step 1. Finding solution

$\frac{\infty }{\infty }$is indeterminate form because infinity is not a real number.

The example may be $\underset{x\to \infty }{\lim }\frac{5x^4-2x}{4x^5-x}=\frac{\infty }{\infty }$

When limit is plug, then it approaches to infinity.

$\underset{x\to \infty }{\lim }\frac{5x^4-2x}{4x^5-x}=\underset{x\to \infty }{\lim }\frac{5-{\displaystyle \frac{2}{x^3}}}{4-{\displaystyle \frac{1}{x^4}}}=\frac{5-2}{4-1}=1$

$\underset{x\to \infty }{\lim }\frac{5x^4+x}{4x^5-3x}=\underset{x\to \infty }{\lim }\frac{5+\frac{1}{x^3}}{4-{\displaystyle \frac{3}{x^4}}}=\frac{5+1}{4-3}=6$

$\underset{x\to \infty }{\lim }\frac{4x^2+x^6}{1-5x^3}=\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{4}{x}}+x^3}{{\displaystyle \frac{1}{x^3}}-5}=\frac{\infty }{-5}=\infty$