Q. 7 - Calculus

Question

Given the following function $f$ , define $f\left(1\right)$ so that $f$ is  continuous at x = 1, if possible:

$f\left(x\right)=\frac{x^2-2x+1}{x^2-6x+5}$

Step-by-Step Solution

Answer

After taking limit 

$\underset{x\to 1}{\lim } \frac{x^2-2x+1}{x^2-6x+5}=0$

Step 1. Given information

Function $f\left(x\right)=\frac{x^2-2x+1}{x^2-6x+5}$

Step 2. Taking limit in the function

$$\begin{gathered} \underset{x\to 1}{\lim } f\left(x\right)=\underset{x\to 1}{\lim } \frac{x^2-2x+1}{x^2-6x+5} \\ =\underset{x\to 1}{\lim } \frac{{(x-1)}^2}{x^2-5x-x+5} \\ =\underset{x\to 1}{\lim } \frac{{(x-1)}^2}{x(x-5)-1(x-5)} \\ =\underset{x\to 1}{\lim } \frac{{(x-1)}^2}{(x-5)(x-1)} \\ =\underset{x\to 1}{\lim } \frac{(x-1)}{(x-5)} \\ =\frac{\underset{x\to 1}{\lim }(x-1)}{\underset{x\to 1}{\lim }(x-5)} \\ =\frac{1-1}{1-5} \\ =\frac{0}{-4} \\ =0 \end{gathered}$$