Question

evaluate limit as x->0 (sinx/x)^(cot(x)^2)

Answer 1

This?\[\Large \lim_{x \rightarrow 0} \left( \dfrac{\sin (x)}{x} \right)^{\cot^2(x)}\]

Answer 2

lim sin(x)
x->0 -------- = 1
x

is the theorem.

Now, cot^2x = cos^2x / sin^2x=1/0=undefined.

So I think the limit doesn't exist.

Answer 3

Limit does exist

Answer 4

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=limit%20as%20x-%3E0%20(sinx%2Fx)%5E(cot(x)%5E2)

Answer 5

Yes, geerky42, that's the one. I know the answer, but I need to actually show the process of getting it and wolfram alpha is saying no step-by-step solution is available.

Answer 6

No idea how to approach it... anything you learned recently that would help us with this problem?

Answer 7

we are working on indeterminate limits such as 0/0, inf/inf, 1^inf, etc. I feel like I have the first few steps right, but it seems there are some trig identities being used and I just can't figure out which one to use when

Answer 8

have you learned taylor series yet?

Answer 9

sin(x)~=x-x^3/6 (taylor series)
(sin(x)/x)^(cot^2(x)) = (1-x^2/6)^(1/sin(x)^2-1)
= (1-x^2/6)^(1/(x-x^3/6)^2-1) = (1-x^2/6)^(1/x^2-1) = (1-x^2/6)^(1/x^2)
let u=-6/x^2. as x->0, u->infinity
= ((1+1/u)^u)^(-1/6)
= e^(-1/6)

Answer 10

we have not done taylor series yet

Answer 11

Did u do it?

Answer 12

no, still haven't figured it out yet

Answer 13

Wait here,for some minutes..

Answer 14

I have tried it about 7 different ways, but no luck yet

Answer 15

Is the answer is 1.

Answer 16

no, e^(-1/6)

Answer 17

what about lhopital? :3

Answer 18

yes, I've used it, but there is much more to it with the exponentital stuff

Answer 19

I haven't. Mind if I give it a shot? Does it give a dead end? ^.^

Answer 20

\[\Large L=\lim_{x \rightarrow 0} \left( \dfrac{\sin (x)}{x} \right)^{\cot^2(x)}\]
\[\Large \ln L=\lim_{x \rightarrow 0} \ \ln\left( \dfrac{\sin (x)}{x} \right)^{\cot^2(x)}\]
\[\Large \ln L=\lim_{x \rightarrow 0} \ [\cot^2(x)]\ln\left( \dfrac{\sin (x)}{x} \right)\]

Answer 21

And then we have (for nothing more than my personal preference :D)
\[\Large \ln L=\lim_{x \rightarrow 0} \left[\frac{\ln\left(\frac{\sin(x)}{x}\right)}{\tan^2(x)}\right]\]

So far, so good, right, @ajnelson23 ? ^.^

Answer 22

yep

Answer 23

can't seem to make the right side equal -1/6

Answer 24

your answers... they make me feel like i'm flying towards a bloody wall.... or at least a wall that will be bloody after I crash into it XD

Answer 25

yep...been working on this for like 2 days looking all over at rules I've never heard of to see if I can make some sense of this

Answer 26

anyway, let's use lhopital on the right side, differentiate both top and bottom of the fraction, we get...
\[\Large \ln L=\lim_{x \rightarrow 0} \left[\frac{x\csc(x)\cdot\frac{x\cos(x)-\sin(x)}{x^2}}{2\sec^2(x)\tan(x)}\right]\]

Answer 27

actually\[\Large \ln L=\lim_{x \rightarrow 0} \left[\frac{\csc(x)\cdot\frac{x\cos(x)-\sin(x)}{x
}}{2\sec^2(x)\tan(x)}\right]\]

Answer 28

yep, got that far

Answer 29

And it looks terrible XD

Answer 30

Let's see...
\[\Large \ln L=\lim_{x \rightarrow 0} \left[\frac{\csc(x)\left[x\cos(x) - \sin(x)\right]}{2x\sec^2(x)\tan(x)}\right]\]

Answer 31

Don't use L'hopital after it ... multiply by x^2 at the top and ... and bottom of x cos(x) - sin(x) / x^3 ... evaluat ethe limits individually.

Answer 32

wait, what?

Answer 33

Still playing with stuff. I'm a bit lost @experimentX :D

\[\Large \ln L=\lim_{x \rightarrow 0} \left[\frac{x\cos(x) - \sin(x)}{2x\sec(x)\tan^2(x)}\right]\]

Answer 34

Just evaluate this limit ..
http://www.wolframalpha.com/input/?i=Limit+x-%3E+0++%28x+cos%28x%29+-+sin%28x%29%29+%2F+x%5E3
rest is easy.

Answer 35

use L'hopital here 3 times ... or you can do it without using L'hopital or power series.

Answer 36

but it get's complicated here :(

Answer 37

I do hate it when I have to L'Hopital more than once :(

Answer 38

so multiply the whole thing by (x^2/x^2) and do l'hospital's again?

Answer 39

yes ... that's what I meant.

Answer 40

@PeterPan you like doing it without using L'hopital ?? there is a way ... just change x -> 3x
but this will get complicated.

Answer 41

how does multiplying (x^2/x^2) times peter pan's limit up there equal the limit you put in wolfram alpha? I am not seeing how to get from one to the other?

Answer 42

hmm what you mean??
the other is just of the form ... just check it
x^2/2 sin(x)^2

Answer 43

that cos thing will go to 1 as x->0

Answer 44

ok, to be more specific, which section above should I multiply times (x^2/x^2)?

Answer 45

\[ \Large \ln L=\lim_{x \rightarrow 0} \left[\frac{x^2\csc(x)}{2\sec^2(x)\tan(x)}\right] \frac{x\cos(x)-\sin(x)}{x^3
} \]
.To get this

Answer 46

Evaluate the left limit and the right limit separately ... on left side, you get 1/2
on right side you get -1/3
hence answer is e^(-1/6)

best way to do this is to use Taylor series as done above.

Answer 47

ok, let me try to work it

Answer 48

the left side is easy ... the right side is hard.

Answer 49

I got!!! Thanks so much!

Answer 50

very good.