What is the limit as x approaches 0 of (tan^2(3x))/(x^2cos(2x)) ?

The limit as x approaches 0 of (tan^2(3x))/(x^2cos(2x)) is 9

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limit as x approaches 0 of (tan^2(3x))/(x^2cos(2x))

x \to 0 lim (\frac{tan ^{2} ( 3 x )}{x ^{2} cos ( 2 x )})

9

Solution steps

x \to 0 lim (\frac{tan ^{2} ( 3 x )}{x ^{2} cos ( 2 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6 sec ^{2} ( 3 x ) tan ( 3 x )}{2 x cos ( 2 x ) - 2 x ^{2} sin ( 2 x )})

= x \to 0 lim (\frac{6 sec ^{2} ( 3 x ) tan ( 3 x )}{2 x cos ( 2 x ) - 2 x ^{2} sin ( 2 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6 ( 6 sec ^{2} ( 3 x ) tan ^{2} ( 3 x ) + 3 sec ^{4} ( 3 x ) )}{2 cos ( 2 x ) - 8 x sin ( 2 x ) - 4 x ^{2} cos ( 2 x )})

= x \to 0 lim (\frac{6 ( 6 sec ^{2} ( 3 x ) tan ^{2} ( 3 x ) + 3 sec ^{4} ( 3 x ) )}{2 cos ( 2 x ) - 8 x sin ( 2 x ) - 4 x ^{2} cos ( 2 x )})

Plug in the value

x = 0

= \frac{6 ( 6 sec ^{2} ( 3 \cdot 0 ) tan ^{2} ( 3 \cdot 0 ) + 3 sec ^{4} ( 3 \cdot 0 ) )}{2 cos ( 2 \cdot 0 ) - 8 \cdot 0 \cdot sin ( 2 \cdot 0 ) - 4 \cdot 0 ^{2} cos ( 2 \cdot 0 )}

Simplify

\frac{6 ( 6 sec ^{2} ( 3 \cdot 0 ) tan ^{2} ( 3 \cdot 0 ) + 3 sec ^{4} ( 3 \cdot 0 ) )}{2 cos ( 2 \cdot 0 ) - 8 \cdot 0 \cdot sin ( 2 \cdot 0 ) - 4 \cdot 0 ^{2} cos ( 2 \cdot 0 )} : 9

= 9

\int_{1}^{2} x^{2} e^{x} d x

\int_{6}^{8} \frac{34}{( x - 6 ) ^{3}} d x

\frac{\partial}{\partial x} (x y (\frac{( 2 - x y )}{x + y}))

\frac{\partial ^{2}}{\partial y \partial x} (\frac{x y}{x - y})

\int \frac{x ^{2} - 36}{x ^{2}} d x

What is the limit as x approaches 0 of (tan^2(3x))/(x^2cos(2x)) ?
The limit as x approaches 0 of (tan^2(3x))/(x^2cos(2x)) is 9