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MATH HOMEWORK

Name: Mark 2006-09-01 13:53

how do i derive lim as x->0 of
(sin5x)/(3x)?

Name: Anonymous 2006-09-01 14:15

Plot it?

Name: Anonymous 2006-09-01 14:17

the teacher wants me to actually solve it by hand..

Name: Anonymous 2006-09-01 15:16

>>1
1/0 > Lim

Name: Anonymous 2006-09-01 15:26

derive top and bottom and oh shi- 5/3!

Name: Anonymous 2006-09-01 17:15

Use L'Hôpital's rule or take the limit of the Maclaurin series of the function as this doesn't produce a indeterminate form.

Name: Anonymous 2006-09-02 0:38

For 0 <= x < pi/2
sin(5x) <= 5x <= tan(5x)
sin(5x)/sin(5x) <= 5x/sin(5x) <= tan(5x)/sin(5x)
1 <= 5x/sin(5x) <= 1/cos(5x)

Lim x->0 of 1 = 1, lim x->0 of 1/cos(5x) = 1, so by Squeeze theorem, lim x->0+ of 5x/sin(5x) is 1.

So, lim x->0+ of sin(5x)/5x = 1, and sin(5x)/3x = 5/3 * sin(5x)/5x, so lim x->0 of sin(5x)/3x = 5/3 * lim x->0+ of sin(5x)/5x = 5/3 * 1 = 5/3.

For -pi/2 < x <= 0
sin(5x) >= 5x >= tan(5x)
(the rest is similar to the above)

Once you have lim x->0+ sin(5x)/3x = 5/3 and lim x->0- sin(5x)/3x = 5/3, you know lim x->0 sin(5x)/3x = 5/3.

Name: Anonymous 2006-09-02 2:20

Don't listen to >>7... he is either slow, or feels the need to show off his maths skills.

>>5 is the way to go.

Name: Anonymous 2006-09-02 3:12

>>7
sin(5x) >= 5x >= tan(5x)*

fix'd

Name: Anonymous 2006-09-02 3:18

l = lim(sin(5x)/3x)
3l = lim(sin(5x)/x)
3/5l = lim(sin(5x)/5x)
3/5l = 1
l = 5/3

Name: Anonymous 2006-09-02 4:42

>>10
3/0=lim(sin(5x)/0)

Name: Anonymous 2006-09-02 4:56

>>11
x=x
x*0=x*0
x=0*x/0
x=0/0

retard.

Name: Anonymous 2006-09-02 6:29

>>12
0/0=x/x

Name: Anonymous 2006-09-03 1:25

>>8
Right, giving a solution that doesn't involve L'Hopital's Theorem (which he likely doesn't know as this is a limit problem that is commonly given near the beginning of a Calc 1 class) makes me slow.

>>9
Not for 0 <= x < pi/2. Example: sin(pi/4) = sqrt(2)/2 = 0.707..., pi/4 = 0.785..., tan(pi/4) = 1. In the other interval, -pi/2 < x <= 0, then you are correct as I said in my post. In short, you fail.

Name: Anonymous 2006-09-04 14:24 (sage)

DO YOUR OWN HOMOWORK FAGGOT

Name: Anonymous 2006-09-06 8:56

>>14
We were taught L'Hopital's at the beginning of calc limits too.

Name: Anonymous 2006-09-06 19:55

>>16
Calculus classes almost always teach limits before the derivative, since that is the way that makes sense. If your teacher did it differently it was a stupid choice on his or her part as it is not possible to do even a halfway decent job of defining the derivative without the notion of a limit.

Name: Anonymous 2006-09-06 21:07

>>17
Limits are gay.

Name: Anonymous 2006-09-08 22:06

I AM UNLIMITED

Name: Anonymous 2006-09-08 23:04

Did you know that cheese comes in a bouqette of many different flavors?

Name: Anonymous 2006-09-09 9:23

CHEESE IS UNLIMITED

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