# Trig limit (strange exponential)

## Homework Statement

$$lim_{x->0} (1+ sin5x)^{cotx}$$

## 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

## The Attempt at a Solution

Mark44
Mentor
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.

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

Mark44
Mentor
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.

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!

Mark44
Mentor
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?

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:
Mark44
Mentor
cot(x) = 1/tan(x)