Question

find the limit as x approaches 0 for cos(2x) / x

Answer 1

(cosxsinx - tanx)/(x^2 sinx)

sinx[cosx - (1/cosx)]/(x^2 sinx)

[cosx - (1/cosx)]/x^2

(cos^2x - 1) /x^2 cosx

- sin^2 x / x^2 cosx = -[sin(x)/x]^2 / cosx

limit x-->0 -[sin(x)/x]^2 / cosx = -(1)^2 / 1 = -1

since we know limit x-->0 sin(x) / x = 1

and cos 0 is 1

Answer 2

O_o

Answer 3

I think there's an easier way to do this . . .

Answer 4

lol

Answer 5

haha. if theres an easier way, please explain.

Answer 6

Well, cosine of whatever ranges between -1 and 1 right?

Answer 7

use sandwich theory..

Answer 8

cos(2x) is varing -1 to 1

Answer 9

and anything between -1 and 1 divided by zero is undefined, so the limit doesn't exist.

Answer 10

i took a mistake no sandwich

Answer 11

I'm going to go eat a sandwich. I'll be back in a couple minutes if you need me.

Answer 12

ugh. trig functions and limits together bug me so much.

Answer 13

use L'hospital theory.

Answer 14

I think L'Hopital's is unreliable here.

Answer 15

why?

Answer 16

Well, it works if you take it far enough, but is more work than is necessary. The reasoning I provided leads to the same conclusion. It's division by zero regardless.

Answer 17

If you only differentiate once, you'll get -2sin(2x)/1 and if you plug in x=0 now you'll mistakenly believe that the limit is zero.
If you keep going and differentiate again, you'll get -4cos(2x)/0 which leads to the same conclusion that I made above.

Answer 18

why it is possible.

Answer 19

it can be convetred when the function combined together

Answer 20

I don't know what you mean.

Answer 21

and L'hosipital theory is appliable right before the denominator is not zero

Answer 22

if certain function f(x) when x goes to 0 assume that its value is infinity and g(x) which is denominator goes to 0 but f/g when x goes to zero, it shoul have certain value

Answer 23

I understand the theory, the limit still does not exist because of division by zero.

Answer 24

ah..
. i had mistake

Answer 25

it is not the 0/0 or infinit/0 or ...

Answer 26

I'm saying that it can be solved with a little logic with no need for calculation.
Remember this, "if theres an easier way, please explain."

Answer 27

it just ... constant/smalllll value

Answer 28

Yup.

Answer 29

so you are right...

Answer 30

it is just infinity....

Answer 31

No, it is not infinity; it is undefined.

Answer 32

......... constant/really small vaue is infinity.. 1/x is ex

Answer 33

If you approach 0 from the left, the left side limit goes to -∞.
Approaching from the right, the right side limit goes to +∞.

Answer 34

Yeah i meant both of two..

Answer 35

okay so if i was going to find the limit as x approaches 0 for sin(3x) / x , would that not exist either?

Answer 36

What if x is a really small negative number? What if x is a really small negative number?
Right, so you can't claim that the limit is infinity.

Answer 37

alright. but i meant that infinity means both of + or -

Answer 38

if numerator is sine, then you can use L'hostpital. it is 0/0

Answer 39

That's fine as long as the final answer is "The limit does not exist."

The limit for sin(3x)/x does exist.

Answer 40

im confused. if you graph cos(2x) / x, the limit is infinity isnt it?

Answer 41

No, if you graph cos(2x)/x the limit at x=0 is undefined, i.e. it does not exist.

Answer 42

and for sin(3x) / x, it would be 3

Answer 43

Yes.

Answer 44

i understand it when i plug it into a graphing calculator. i just dont understand how to do it algebraically.

Answer 45

LeeYeong's suggestion of L'Hopital's rule is appropriate here.

Answer 46

One way to think of it is to imagine the triangle that represents the sin(x)/x situation.