L'Hopital's Rule Solution 6

SOLUTION 6: $$ \displaystyle{ \lim_{x \rightarrow 0} \frac{ x \tan x }{ \sin 3x } } = \displaystyle{ \frac{ 0 \cdot \tan(0) }{ \sin(0) } } = \displaystyle{ \frac{ 0 \cdot 0 }{ 0 } } = \frac{"0"}{0} $$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately. Use the Product Rule on the top. Use the Chain Rule on the bottom.)
$$ = \displaystyle{ \lim_{x \rightarrow 0} \frac{ x \sec^2 x + (1) \tan x }{ 3 \cos x } } $$ $$ = \displaystyle{ \frac{ (0) \sec^2(0) + \tan(0) }{ 3 \cos(0) } } $$ $$ = \displaystyle{ \frac{ (0)(1)^2 + 0 }{ 3 (1) } } $$ $$ = \displaystyle{ \frac{ 0 }{ 3 } } $$ $$ = \displaystyle{ 0 } $$

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