Question

How do you find the limit as x approaches `pi/4` of `[sin(x) - cos(x)] / cos(2x)`?

Answer 1

`sqrt2/(-2)`

This is an indeterminate form of the type `0/0`, hence L'Hopital's rule would apply and limit can be evaluated by differentiating numerator and denominator and then applying the limit. Accordingly,

Lim `x->pi/4` `(sinx -cosx)/cos(2x)`

= Lim`x->pi/4` `(cosx +sinx)/(-2sin2x)`

= `sqrt2/(-2)`

Answer 2

If you want to evaluate the limit without l'Hopital's Rule (find the limit "algebraically"), do this:

Use
`cos2x = cos^2x-sin^2x = (cosx-sinx)(cosx+sinx)`

`lim_(xrarr pi/4) (sinx-cosx)/cos(2x) = lim_(xrarr pi/4) (- 1(cosx-sinx))/((cosx-sinx)(cosx+sinx))`

`=lim_(xrarr pi/4) (-1)/(cosx+sinx) = (-1)/(1/sqrt2+1/sqrt2)=(-1)/(2/sqrt2) = -sqrt2/2`