Question

Question on limits... to be posted below.

Answer 1

It is known that \[\lim_{x \rightarrow \infty} (1+ \frac{ 2 }{ x })^x = e^2\] use this fact, other information about basic limits, and some appropriate algebra, to show that \[\lim_{y \rightarrow 0^+} (1+siny)^\frac{ 2 }{ y }\]

Answer 2

Try applying log on both sides putting an x as the result of the limit.

Answer 3

on which limit?

Answer 4

The one you want to know.

Answer 5

What I was thinking when I saw that, was that the limit of (sin y)/y=1 when y tends to zero, and the only way of putting that y of the exponent down is applying log

Answer 6

hmmmm

Answer 7

I think the answer is 1 but I'm not sure.

Answer 8

if the limit is approaching 0 from the positive side

Answer 9

then for the 2/y, you would get 2 over a very small positive number, this is thought to be infinity

Answer 10

\[\lim_{y \rightarrow 0^+} (1+siny)^\frac{ 2 }{ y }= e^2\]

Answer 11

yep, take the ln

Answer 12

what he said.

Answer 13

that would bring the 2/y down and make it just 2 on the other side.

Answer 14

I'm not getting any luck with that

Answer 15

\[\lim_{y \rightarrow 0^+} \ln[(1+siny)^\frac{ 2 }{ y }] =2\]
\[\lim_{y \rightarrow 0^+} \ \frac{ 2}{ y } \ln(1+siny) =2\]

Answer 16

Where did you get that 2 from? On the right hand side

Answer 17

ln(e^2)

Answer 18

But we don't know that this limit is e^2

Answer 19

You cant use the answer

Answer 20

okay.. sorry.. anyways \[\lim_{y \rightarrow 0^+} \ \frac{ 2}{ y } \ln(1+siny) \]

Answer 21

is there an e^2 there? couse I still cant see it

Answer 22

Or this was for nasryn? Do we know the answer or not?

Answer 23

I don't know what the answer is, but i'm pretty sure this limit = e^2 from what my graphing calc told me

Answer 24

Ok, so we cant use it to answer.

Answer 25

ok, then how?

Answer 26

Do you know l'hospital rules?

Answer 27

Because I found a way of doing using it

Answer 28

no i don't, it's not needed apparently

Answer 29

@Hero maybe you can shine some light?

Answer 30

Are you sure the original question is complete as posted? Seems like something is missing. Should that second equation be set to e^2 or did you just add that in there?

Answer 31

That e^2 was added in later. the original question is correct at the top

Answer 32

You cannot apply logs then because you would only be applying it to one side.

Answer 33

Good call.

Answer 34

The next question is: Clearly identify the substitution that is needed for the solution.

Answer 35

We use the fact that \[\sin x=\frac{ e^{ix}-e ^{-ix} }{ 2 }\]

Answer 36

thats sinhx

Answer 37

2i in the denominator sorry

Answer 38

oh sorry, didn't see the i's, but we haven't even dealt with complex numbers like that

Answer 39

Whatever the substitution is, it would have to allow for the insertion of y = 0 into the expression.

Answer 40

Ok forget it then. If they give us that formula for e^2 we shoul try to transform it into somthing like that.

Answer 41

well, the thing is, it is saying insert y = 0 from the right, so it is best to insert small positive values of y between 0 < y < 0.1

Answer 42

Combine that with graphing the expression and you should get your answer.

Answer 43

I think it whould have a way of solving it analyticaly

Answer 44

I don't think there is a way around graphing this and the clue for this is that you can only consider small values of y close to zero but only for the right.

Answer 45

as I said, with l'osital we can get it, but since he has not studied that we cant use it but it is analyticaly possible