Q. 50 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to -\infty }{\lim }\frac{1-2x^2}{(3-x)(3+4x)}$

Step-by-Step Solution

Answer

The limit is $\frac{1}{2}$

Step 1. Given Information:

Given expression: $\underset{x\to -\infty }{\lim }\frac{1-2x^2}{(3-x)(3+4x)}$

We want to find limit of given expression.

Step 2. Solution:

Simplify the denominator we get

$\underset{x\to -\infty }{\lim }\frac{1-2x^2}{(3-x)(3+4x)}=\underset{x\to -\infty }{\lim }\frac{1-2x^2}{9+9x-4x^2}$

Now take x² common form numerator and denominator and simplify we get

$\underset{x\to -\infty }{\lim }\frac{1-2x^2}{9+9x-4x^2}=\underset{x\to -\infty }{\lim }\frac{x^2({\displaystyle \frac{1}{x^2}}-2)}{x^2({\displaystyle \frac{9}{x^2}}+{\displaystyle \frac{9}{x}}-4)}=\underset{x\to -\infty }{\lim }\frac{{\displaystyle \frac{1}{x^2}}-2}{{\displaystyle \frac{9}{x^2}}+{\displaystyle \frac{9}{x}}-4}$

Now take limit we get

$\underset{x\to -\infty }{\lim }\frac{{\displaystyle \frac{1}{-{\infty }^2}}-2}{{\displaystyle \frac{9}{-{\infty }^2}}+{\displaystyle \frac{9}{\infty }}-4}=\underset{x\to -\infty }{\lim }\frac{-2}{-4}=\underset{x\to -\infty }{\lim }\frac{4}{2}=\frac{1}{2}$