Q. 15 - 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.

15. $\underset{x\to 1}{\lim }\frac{x^2+x-2}{x-1}$

Step-by-Step Solution

Answer

The value of the limit is $3$

Part (a) Step 1. Given information

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

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

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{x^2+x-2}{x-1} \\ =\underset{x\to 1}{\lim }\frac{2x+1}{1} \\ =2\left(1\right)+1 \\ =3 \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{x^2+x-2}{x-1} \\ =\underset{x\to 1}{\lim }\frac{x^2+2x-x-2}{x-1} \\ =\underset{x\to 1}{\lim }\frac{x\left(x+2\right)-1\left(x+2\right)}{x-1} \\ =\underset{x\to 1}{\lim }\frac{\left(x+2\right)\left(x-1\right)}{x-1} \\ =\underset{x\to 1}{\lim }x+2 \\ =1+2 \\ =3 \end{gathered}$$