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Trig limit (strange exponential)

  • Thread starter holezch
  • Start date
  • #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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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4,854
An obvious approach in this type of problem is to let y = (1 + sin(5x))cot x, and then take the natural log of both sides.

Look in your textbook and see if you can find some examples that use this technique.
 
  • #3
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okay thanks, I've never seen something like this before.
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
33,169
4,854
Yes, use L'Hopital's Rule.

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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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?

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
33,169
4,854
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?
Here's the situation. You have y = (1 + sin5x)cotx. Taking the natural log of both sides gives:
ln y = ln (1 + sin5x)cotx

Now take the limit as x -> 0 of both sides:
lim ln y = lim ln (1 + sin5x)cotx

As long as the functions involved are continuous you can interchange the lim and ln operations, yielding:
ln (lim y) = lim ln (1 + sin5x)cotx

What you're interested in is lim y = lim ln (1 + sin5x)cotx. What you'll be getting above is the log of lim y.
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.
Where are you getting cos(x)? The exponent is cot(x), not cos(x).
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!
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?
 
  • #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
 
Last edited:
  • #8
33,169
4,854
cot(x) = 1/tan(x)
 

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