What is the limit as x approaches pi/4 of (sec^2(x)-2tan(x))/(1+cos(4x)) ?

The limit as x approaches pi/4 of (sec^2(x)-2tan(x))/(1+cos(4x)) is 1/2

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limit as x approaches pi/4 of (sec^2(x)-2tan(x))/(1+cos(4x))

x \to \frac{π}{4} lim (\frac{sec ^{2} ( x ) - 2 tan ( x )}{1 + cos ( 4 x )})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

x \to \frac{π}{4} lim (\frac{sec ^{2} ( x ) - 2 tan ( x )}{1 + cos ( 4 x )})

Apply L'Hopital's Rule

: x \to \frac{π}{4} lim (\frac{2 sec ^{2} ( x ) tan ( x ) - 2 sec ^{2} ( x )}{- 4 sin ( 4 x )})

= x \to \frac{π}{4} lim (\frac{2 sec ^{2} ( x ) tan ( x ) - 2 sec ^{2} ( x )}{- 4 sin ( 4 x )})

Apply L'Hopital's Rule

: x \to \frac{π}{4} lim (\frac{- 4 sec ^{2} ( x ) + 6 sec ^{4} ( x ) - 4 sec ^{2} ( x ) tan ( x )}{- 16 cos ( 4 x )})

= x \to \frac{π}{4} lim (\frac{- 4 sec ^{2} ( x ) + 6 sec ^{4} ( x ) - 4 sec ^{2} ( x ) tan ( x )}{- 16 cos ( 4 x )})

Plug in the value

x = \frac{π}{4}

= \frac{- 4 sec ^{2} ( \frac{π}{4} ) + 6 sec ^{4} ( \frac{π}{4} ) - 4 sec ^{2} ( \frac{π}{4} ) tan ( \frac{π}{4} )}{- 16 cos ( 4 \cdot \frac{π}{4} )}

Simplify

\frac{- 4 sec ^{2} ( \frac{π}{4} ) + 6 sec ^{4} ( \frac{π}{4} ) - 4 sec ^{2} ( \frac{π}{4} ) tan ( \frac{π}{4} )}{- 16 cos ( 4 \cdot \frac{π}{4} )} : \frac{1}{2}

= \frac{1}{2}

y^{^{''}} - (\frac{6}{x}) y^{^{'}} - (\frac{9}{x ^{2}}) y = 0

x \to - 2 + lim (\frac{- 2}{x ^{2} - 4})

x \to + 0 lim (\frac{1}{x})

x^{^{''}} + k x^{^{'}} + x = 0

\frac{d}{d x} (\frac{3 x ^{2}}{4})

What is the limit as x approaches pi/4 of (sec^2(x)-2tan(x))/(1+cos(4x)) ?
The limit as x approaches pi/4 of (sec^2(x)-2tan(x))/(1+cos(4x)) is 1/2