- #1

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## Homework Statement

[tex] lim_{x->0} (1+ sin5x)^{cotx} [/tex]

## Homework Equations

that's the problem.. I don't know :s

## The Attempt at a Solution

can't think of any theorems or any methods I could use here.. what should I do? thank you

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- Thread starter holezch
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- #1

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[tex] lim_{x->0} (1+ sin5x)^{cotx} [/tex]

that's the problem.. I don't know :s

can't think of any theorems or any methods I could use here.. what should I do? thank you

- #2

Mark44

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Look in your textbook and see if you can find some examples that use this technique.

- #3

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when I do that, I'll get lim x-> 0 cotxln(1+sinx) = lny, ln(1+sinx) looks alright, since sinx goes to 0 as x goes to 0, but what should I do about cotx? should I express this in another way to get 0/0 and use l'hopitals?

thank you

- #4

Mark44

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Also, you'll have lim x -> 0 ln y = lim x -> cot x * ln(1 + sin 5x). For the expression on the left, the standard technique is to reverse the order of the limit and log operations, to get ln(lim x ->0 y) = ...

Keep in mind that you'll be getting the ln of your limit.

- #5

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also, what kind of identity should I use for my cos x to get my 0/0? I tried sqrt(1-sin^2x) and playing around with the difference of squares, but it didn't work so well.

I feel like I might run into trouble with my ln(1+sinx) also, since that equals to zero, the product still might not go to zero, but I'll have to find a way around it. :S

thanks for the help!

- #6

Mark44

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Here's the situation. You have y = (1 + sin5x)thanks for the reply, I don't really understand the meaning behind doing ln(lim x -> 0 y) versus lim x-> 0 ln(y) when I'll just ln( y) either way?

ln y = ln (1 + sin5x)

Now take the limit as x -> 0 of both sides:

lim ln y = lim ln (1 + sin5x)

As long as the functions involved are continuous you can interchange the lim and ln operations, yielding:

ln (lim y) = lim ln (1 + sin5x)

What you're interested in is lim y = lim ln (1 + sin5x)

Where are you getting cos(x)? The exponent is cot(x), not cos(x).also, what kind of identity should I use for my cos x to get my 0/0? I tried sqrt(1-sin^2x) and playing around with the difference of squares, but it didn't work so well.

You're being very sloppy. This is the second time you have mentioned ln(1 + sinx). It's ln(1 + sin(5x)). The product is cot(x)*ln(1 + sin5x). Can you think of a way to write this as a quotient suitable for use in L'Hopital's Rule, rather than a product?I feel like I might run into trouble with my ln(1+sinx) also, since that equals to zero, the product still might not go to zero, but I'll have to find a way around it. :S

thanks for the help!

- #7

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I was looking at cosx because we have cot(x ) , so we have cos(x)/sin(x) and I want to get 0/0 from this.

By the way, sorry about the 1+sinx v.s. 1+sin 5x, I just made a typo

but I see now that we're taking the limit as a whole, and not by separate products

thanks, I'll try

By the way, sorry about the 1+sinx v.s. 1+sin 5x, I just made a typo

but I see now that we're taking the limit as a whole, and not by separate products

thanks, I'll try

Last edited:

- #8

Mark44

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cot(x) = 1/tan(x)

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