# could you please calculate the following limit? lim┬(x→π/4)〖(〖cos〗^4 x-〖sin〗^4 x)/sin4x〗 thanks

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Calculate `lim_(x->pi/4)(cos^4x-sin^4x)/(sin4x)` :

Simplify the argument:

`cos^4x-sin^4x=(cos^2x-sin^2x)(cos^2x+sin^2x)`

Use the Pythagorean relationship `cos^2x+sin^2x=1` so

`cos^4x-sin^4x=cos^2x-sin^2x`

Also `sin4x=2sin2xcos2x=4sinxcosxcos2x`

(Using `sin2u=2sinucosu` )

Now use `cos2u=cos^2u-sin^2u` to get:

`sin4x=4sinxcosx(cos^2x-sin^2x)`

So `(cos^4x-sin^4x)/(sin4x)=(cos^2x-sin^2x)/(4sinxcosx(cos^2x-sin^2x))`

`=1/(4sinxcosx)`

Now we use the fact that if two functions f(x) and g(x) agree at every point (except possible at c), and if lim g(x) exists then lim f(x)=lim g(x):

Therefore `lim_(x->pi/4)(cos^4x-sin^4x)/(sin4x)`

`=lim_(x->pi/4)1/(4sinxcosx)` We can use substitution to get:

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`lim_(x->pi/4)(cos^4x-sin^4x)/(sin4x)=1/2`

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