Q. 16 - 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.

16. $\underset{x\to 2}{\lim }\frac{x^2-4x+4}{x-2}$

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 $\underset{x\to 2}{\lim }\frac{x^2-4x+4}{x-2}$

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

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^2-4x+4}{x-2} \\ =\underset{x\to 2}{\lim }\frac{2x-4}{1} \\ =2\left(2\right)-4 \\ =0 \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^2-4x+4}{x-2} \\ =\underset{x\to 2}{\lim }\frac{{\left(x-2\right)}^2}{x-2} \\ =\underset{x\to 2}{\lim }x-2 \\ =2-2 \\ =0 \end{gathered}$$