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March 5, 2015

March 5, 2015

Posted by **Alex** on Saturday, February 21, 2009 at 4:39pm.

The (x-->pi/4) goes below the lim.

Should i just substitute the x's with pi/4??

- Calculus -
**Alex**, Saturday, February 21, 2009 at 5:07pmThis question is related to limits.

- Calculus -
**drwls**, Saturday, February 21, 2009 at 5:15pmIf 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 -
**drwls**, Saturday, February 21, 2009 at 5:18pmThat'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 -
**bobpursley**, Saturday, February 21, 2009 at 5:20pmif 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 -
**bobpursley**, Saturday, February 21, 2009 at 5:22pmlooking 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 -
**Rina**, Saturday, February 21, 2009 at 5:33pmThanks. 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 -
**Alex**, Saturday, February 21, 2009 at 5:39pmThanks everyone!

- Calculus -
**Alex**, Saturday, February 21, 2009 at 5:48pmI 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

Is this answer right??

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