L'Hopital's Rule Solution 5

SOLUTION 5: $$ \displaystyle{ \lim_{x \rightarrow 3} \frac{ 1/x - 1/3 }{ x^2-9 } } = \displaystyle{ \frac{ 1/(3) - 1/3 }{ (3)^2-9 } } = \displaystyle{ \frac{ 0 }{ 9-9 } } = \frac{"0"}{0} $$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately.)
$$ = \displaystyle{ \lim_{x \rightarrow 3} \frac{ -1/x^2 - 0 }{ 2x-0 } } $$ $$ = \displaystyle{ \lim_{x \rightarrow 3} \Big( \frac{ -1 }{ x^2 } \cdot \frac{ 1 }{ 2x } \Big) } $$ $$ = \displaystyle{ \lim_{x \rightarrow 3} \frac{ -1 }{ 2x^3 } } $$ $$ = \displaystyle{ \frac{ -1 }{ 2(3)^3 } } $$ $$ = \displaystyle{ \frac{ -1 }{ 54 } } $$

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