Answer:
[tex]Lim_{x \to 5}f(x)=f(5)[/tex].
Step-by-step explanation:
The given function is
[tex]f(x)=\frac{x^2-49}{(x^2+6x-7)(x^2+14x+49)}[/tex].
We can factor an rewrite for easy evaluation;
[tex]f(x)=\frac{(x-7)(x+7)}{(x-1)(x+7)(x+7)^2}[/tex].
[tex]f(x)=\frac{(x-7)}{(x-1)(x+7)^2}[/tex].
Let us plug in 5.
[tex]f(5)=\frac{(5-7)}{(5-1)(5+7)^2}[/tex].
[tex]f(5)=\frac{-2}{576}[/tex].
[tex]f(5)=-\frac{1}{288}[/tex].
Let us find the Left Hand Limit;
[tex]Lim_{x \to 5^-}f(x)=\frac{(5-7)}{(5-1)(5+7)^2}=-\frac{1}{288}[/tex].
Now the Right Hand Limit;
[tex]Lim_{x \to 5^+}f(x)=\frac{(5-7)}{(5-1)(5+7)^2}=-\frac{1}{288}[/tex].
Since the One-Sided Limits exist and are equal;
[tex]Lim_{x \to 5}f(x)=\frac{(5-7)}{(5-1)(5+7)^2}=-\frac{1}{288}[/tex].
We have shown that;
[tex]Lim_{x \to 5}f(x)=f(5)[/tex].
This is the definition of continuity.
Hence f(x) is continuous at x=5
