Answer:
The function is continuous at x=3
Step-by-step explanation:
Continuity of a Function
There are three conditions a function f(x) must comply to be continuous at x=a:
- [tex]\lim\limits_{x \rightarrow a} f(x)[/tex] must exist
- [tex]\lim\limits_{x \rightarrow a} f(x)=f(a)[/tex]
Should one of the conditions fail, the function would not be continuous at x=a.
Let's test the continuity of the function at x=3
[tex]\displaystyle\left\{\begin{matrix}\frac{x^2-4x+3}{x-3} & \text{if }x\neq 3 \\ 2 & \text{if }x= 3\end{matrix}\right.[/tex]
First, we find out if f(3) exists. The second condition of the piecewise function indicates that f(3) = 2
Now let's compute
[tex]\displaystyle \lim\limits_{x \rightarrow 3} \frac{x^2-4x+3}{x-3}[/tex]
This is an indeterminate limit which must be simplified by factoring the numerator:
[tex]\displaystyle \lim\limits_{x \rightarrow 3} \frac{(x-3)(x-1)}{x-3}[/tex]
Simplifying
[tex]\displaystyle \lim\limits_{x \rightarrow 3} (x-1)=2[/tex]
The limit exists and is equal to 2
The third condition demands that the function and the limit are equal. We can see this condition is met, thus we conclude the function is continuous at x=3
