Use the definition of continuity and the properties of limits to show that (Picture Provided Below)

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bm42400

High School

Use the definition of continuity and the properties of limits to show that (Picture Provided Below)

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