Q. 49 - Calculus

Question

Calculate each limit in Exercises 35–80.  

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

Step-by-Step Solution

Answer

The limit is - 9

Step 1. Given Information

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

We want to find limit of given expression. 

Step 2. Solution:

Simplify:

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

We know that $a^3-b^3=(a-b)(a^2+b^2+ab) and {(a+b)}^2=a^2+2ab+b^2$

So we have

$\underset{x\to \infty }{\lim }\frac{(9x^2+6x+1)(x-1)}{(1-x)(1+x^2+x)}=\underset{x\to \infty }{\lim }-\frac{(9x^2+6x+1)}{(1+x^2+x)}$

Take x² Common from numerator and denominator we get

$\underset{x\to \infty }{\lim }-\frac{x^2(9+{\displaystyle \frac{6}{x}}+{\displaystyle \frac{1}{x^2}})}{x^2({\displaystyle \frac{1}{x^2}}+1+{\displaystyle \frac{1}{x}})}=\underset{x\to \infty }{\lim }-\frac{9+{\displaystyle \frac{6}{x}}+{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}+1+{\displaystyle \frac{1}{x}}}$

Now take limit we get:

$\underset{x\to \infty }{\lim }-\frac{9+{\displaystyle \frac{6}{x}}+{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}+1+\frac{1}{x}}}=-\frac{9+{\displaystyle \frac{6}{\infty }}+{\displaystyle \frac{1}{{\infty }^2}}}{{\displaystyle \frac{1}{{\infty }^2}+1+\frac{1}{\infty }}}=-\frac{9}{1}=-9$