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Is there a way to define $f(c)$ for the for the following function so that $f(x)$ is continuous at $x=\mathrm{c}$ ?
$f(x)=\dfrac{x^{2}-16}{x-4}$ at $c=4$
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## 1 Answer

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$\lim _{x_{x}+4} f(x)=\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}$

$=\lim _{x \rightarrow 4}\left(\frac{(x+4)(x-4)}{x-4}\right)$

$=\lim _{x \rightarrow 4}(x+4) =8$

Because $f(x)=8$, we should define $\lim _{x \rightarrow 4} f(4)=8$ to make this function continuous.
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