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# Evaluate the limit with either L’Hoopital’s rule or previously learned methods.$\lim _{x \rightarrow 3} \frac{x^{2}-9}{x+3}$

Solution

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Since $f ( x ) = \frac { x ^ { 2 } - 9 } { x + 3 }$ is defined for all $x \not =-3$ we can evaluate the limit directly as:

\begin{aligned} \lim _ { x \rightarrow 3 } \frac { x ^ { 2 } - 9 } { x + 3 } = \frac { ( 3 ) ^ { 2 } - 9 } { 3 + 3 } = \frac { 0 } { 6 } = 0 \end{aligned}

Finally, we have found that:

\begin{aligned} \lim _ { x \rightarrow 3 } \frac { x ^ { 2 } - 9 } { x + 3 } = 0 \end{aligned}

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