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Question

Evaluate the limit with either L’Hoopital’s rule or previously learned methods.

limx3x29x+3\lim _{x \rightarrow 3} \frac{x^{2}-9}{x+3}

Solution

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Since f(x)=x29x+3f ( x ) = \frac { x ^ { 2 } - 9 } { x + 3 } is defined for all x3x \not =-3 we can evaluate the limit directly as:

limx3x29x+3=(3)293+3=06=0\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:

limx3x29x+3=0\begin{aligned} \lim _ { x \rightarrow 3 } \frac { x ^ { 2 } - 9 } { x + 3 } = 0 \end{aligned}

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