# Chapter 3 - The Derivative - Chapter Review - Review Exercises - Page 189: 40

$$f(x)=\frac{x^{2}-9}{3+x}$$ $x$-value where the given function is discontinuous. is $$x=-3$$ and $$f(-3)$$ do not exist, also $$\lim _{x \rightarrow -3} f(x)$$ is exist and equal $-6$

#### Work Step by Step

$$f(x)=\frac{x^{2}-9}{3+x}$$ This function is discontinuous wherever the denominator is zero. There is a discontinuity when $x=-3$ . so the given function is discontinuous at $x=-3$, therefore $f(-3)$ do not exist. Now, we investigate $\lim _{x \rightarrow -3} f(x)$ at $x=-3$ $$\begin{split} \lim _{x \rightarrow -3} f(x) &=\lim _{x \rightarrow -3}\left(\frac{x^{2}-9}{3+x} \right) \\ &=\lim _{x \rightarrow -3}\left(\frac{(x-3)(x+3)}{(3+x)} \right)\\ &=\lim _{x \rightarrow -3}\left(x-3 \right)\\ &=-3-3 \\ &=-6 \end{split}$$ so, $$\lim _{x \rightarrow -3} f(x)$$ is exist and equal $-6$. Hence, $x$-value where the given function is discontinuous. is $$x=-3$$ and $$f(-3)$$ do not exist, also $$\lim _{x \rightarrow -3} f(x)$$ is exist and equal $-6$.

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