Q. 17 - Calculus

Question

Calculate each of the limits in Exercises 15–20 (a) using L’Hôpital’s rule and (b) without using L’Hôpital’s rule.

17. $\underset{x\to \infty }{\lim }\frac{x-1}{2-3x^2}$

Step-by-Step Solution

Answer

The value of the limit is $0$

Part (a) Step 1. Given information

We have been given the limit as $\underset{x\to \infty }{\lim }\frac{x-1}{2-3x^2}$

Part (a) Step 2. Find the limit using L’Hôpital’s rule.

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x-1}{2-3x^2} \\ =\underset{x\to \infty }{\lim }\frac{1}{-6x} \\ =0 \end{gathered}$$

Part (b) Step 1. Find the limit without using L’Hôpital’s rule.

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x-1}{2-3x^2} \\ =\underset{x\to \infty }{\lim }\frac{{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{2}{x^2}}-3} \\ =\underset{x\to \infty }{\lim }\frac{0-0}{0-3} \\ =\frac{0}{-3} \\ =0 \end{gathered}$$