Calculus
posted by Alex on .
How do I solve lim(x>pi/4)(sinxcosx)/cos(2x)
The (x>pi/4) goes below the lim.
Should i just substitute the x's with pi/4??

This question is related to limits.

If you do that, you will get 0/0. You have to take the ratio of the derivatives.
lim(x>pi/4)(sinxcosx)/cos(2x)
= lim(x>pi/4)(cosx+sinx)/2 sin(2x)
= sqrt 2/(2) = 0.707 
That's called using L'Hopital's rule for indeterminateratio limits. It works for 0/0 or infinity/infinity. Try it with a number close to pi/4, such as 0.785, and you will see that it gives the right answer.

if you put in PI/4 in to the x, you will get an indeterminate form 0/0.
Lim (sinx/cos2x  cosx/cos2x)
multiply numerator and denominator by (sinx+cosx)
lim (sin^2x cos^2x)/cos2x(sinx+cosx)
lim (sin^2xcos^2x)/(sin^2xcos^2x)(sinx+cosx)
lim 1/(sinx+cosx)= 1/sqrt2
check that. 
looking at drwls answer, using L'Hopitals rule (I assumed you had not gotten to it yet), I see I must have made a sign error, you can find it.

Thanks. Actually i did not learn that other rule yet and I don't think we will but thanks anyway it was helpful.
I solved using bobpursley's idea and got sqrt2/2=1/sqrt2. 
Thanks everyone!

I have a similar question relating to limits that I solved
lim(x>0) (1cosx)/2x^2
I multiplied the numerator and denominator by (1+cosx) to get
lim(x>0) (1cos^2x)/2x^2(1+cosx)
= lim(x>0)sin^2x/2x^2(1+cosx)
the lim(x>0) (sinx/x)^2 would =1
I then substituted the remaining portion with 0 and got 1/4
Is this answer right??