What is the limit as x approaches 0 of (tan^2(4x))/(1-cos(5x)) ?

The limit as x approaches 0 of (tan^2(4x))/(1-cos(5x)) is 32/25

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limit as x approaches 0 of (tan^2(4x))/(1-cos(5x))

x \to 0 lim (\frac{tan ^{2} ( 4 x )}{1 - cos ( 5 x )})

\frac{32}{25}

Decimal

1.28

Solution steps

x \to 0 lim (\frac{tan ^{2} ( 4 x )}{1 - cos ( 5 x )})

Apply L'Hopital's Rule

= x \to 0 lim (\frac{8 sec ^{2} ( 4 x ) tan ( 4 x )}{5 sin ( 5 x )})

Apply L'Hopital's Rule

= x \to 0 lim (\frac{8 ( 8 sec ^{2} ( 4 x ) tan ^{2} ( 4 x ) + 4 sec ^{4} ( 4 x ) )}{25 cos ( 5 x )})

Plug in the value

x = 0

= \frac{8 ( 8 sec ^{2} ( 4 \cdot 0 ) tan ^{2} ( 4 \cdot 0 ) + 4 sec ^{4} ( 4 \cdot 0 ) )}{25 cos ( 5 \cdot 0 )}

Simplify

\frac{8 ( 8 sec ^{2} ( 4 \cdot 0 ) tan ^{2} ( 4 \cdot 0 ) + 4 sec ^{4} ( 4 \cdot 0 ) )}{25 cos ( 5 \cdot 0 )} : \frac{32}{25}

= \frac{32}{25}

What is the limit as x approaches 0 of (tan^2(4x))/(1-cos(5x)) ?
The limit as x approaches 0 of (tan^2(4x))/(1-cos(5x)) is 32/25