# Calculus

posted by on .

How do I solve lim(x-->pi/4)(sinx-cosx)/cos(2x)
The (x-->pi/4) goes below the lim.
Should i just substitute the x's with pi/4??

• Calculus - ,

This question is related to limits.

• Calculus - ,

If you do that, you will get 0/0. You have to take the ratio of the derivatives.

lim(x->pi/4)(sinx-cosx)/cos(2x)
= lim(x->pi/4)(cosx+sinx)/-2 sin(2x)
= sqrt 2/(-2) = -0.707

• Calculus - ,

That's called using L'Hopital's rule for indeterminate-ratio 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.

• Calculus - ,

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^2x-cos^2x)/(sin^2x-cos^2x)(sinx+cosx)

lim 1/(sinx+cosx)= 1/sqrt2

check that.

• Calculus - ,

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.

• Calculus - ,

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.

• Calculus - ,

Thanks everyone!

• Calculus - ,

I have a similar question relating to limits that I solved
lim(x-->0) (1-cosx)/2x^2
I multiplied the numerator and denominator by (1+cosx) to get
lim(x->0) (1-cos^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