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

Question

Emma Aldridge · Accepted Answer

#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)#

Christopher Allers · Answer

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#

Cameron Alexander · Answer

undefined To find the limit as x approaches pi/4 of [sin(x) - cos(x)] / cos(2x), we can use algebraic manipulation and trigonometric identities.

First, we simplify the expression by expanding cos(2x) using the double angle formula: cos(2x) = cos^2(x) - sin^2(x).

Next, we substitute this expression back into the original equation: [sin(x) - cos(x)] / (cos^2(x) - sin^2(x)).

Now, we can factor out a common factor of (sin(x) - cos(x)) from the numerator: (sin(x) - cos(x)) / [(cos(x) + sin(x))(cos(x) - sin(x))].

Since sin(x) - cos(x) is a common factor in both the numerator and denominator, we can cancel it out: 1 / (cos(x) + sin(x)).

Finally, we substitute x = pi/4 into the expression: 1 / (cos(pi/4) + sin(pi/4)).

Evaluating cos(pi/4) and sin(pi/4), we get: 1 / (√2/2 + √2/2) = 1 / √2 = √2/2.

Therefore, the limit as x approaches pi/4 of [sin(x) - cos(x)] / cos(2x) is √2/2.