# Limit of arctan(e^x)?

(sorry the thread title is wrong - can a mod please change it to "Limit of e^-7x cos x?")
1. Homework Statement

Find the following:
$$\lim_{x \rightarrow \infty} e^{-7x} \cos x$$

## Homework Equations

I know that $$[ \lim_{x \rightarrow a} f(x)g(x) ] = [ \lim_{x \rightarrow a} f(x) ] \cdot [ \lim_{x \rightarrow a} g(x)]$$ so...

## The Attempt at a Solution

From the above equations...
$$[\lim_{x \rightarrow \infty} e^{-7x} \sin x] = [(\lim_{x \rightarrow \infty} e^{-7x}) \cdot (\lim_{x \rightarrow \infty} \cos x)]$$ But this limit DNE, since $$\lim_{x \rightarrow \infty} cos x : DNE$$

And yet, WebAssign tells me that DNE is wrong, and that I must use the Squeeze Theorem for this problem. Any help? My trig is a little rusty (last time I took it was 9 years ago), so any useful reminders on that front would be helpful as well. I'm taking Calc I.

Dick
Homework Helper
(sorry the thread title is wrong - can a mod please change it to "Limit of e^-7x cos x?")
1. Homework Statement

Find the following:
$$\lim_{x \rightarrow \infty} e^{-7x} \cos x$$

## Homework Equations

I know that $$[ \lim_{x \rightarrow a} f(x)g(x) ] = [ \lim_{x \rightarrow a} f(x) ] \cdot [ \lim_{x \rightarrow a} g(x)]$$ so...

## The Attempt at a Solution

From the above equations...
$$[\lim_{x \rightarrow \infty} e^{-7x} \sin x] = [(\lim_{x \rightarrow \infty} e^{-7x}) \cdot (\lim_{x \rightarrow \infty} \cos x)]$$ But this limit DNE, since $$\lim_{x \rightarrow \infty} cos x : DNE$$

And yet, WebAssign tells me that DNE is wrong, and that I must use the Squeeze Theorem for this problem. Any help? My trig is a little rusty (last time I took it was 9 years ago), so any useful reminders on that front would be helpful as well. I'm taking Calc I.

Just because one of the limits DNE doesn't mean that the product DNE. Could it be that one of your limits is 0? Use the squeeze theorem!

Just because one of the limits DNE doesn't mean that the product DNE. Could it be that one of your limits is 0? Use the squeeze theorem!

Okay, I can see that $$\lim_{x \rightarrow \infty} e^{-7x} = 0$$ Which means that the limit as a whole should be 0, if it's not that case that the limit DNE. In that case, can I use the Squeeze Theorem like this?
$$[-1 \leq \cos x \leq 1] \Rightarrow [-e^{-7x} \leq e^{-7x} cos x \leq e^{-7x}]$$ Meaning that $$[\lim_{x \rightarrow \infty}(-e^{-7x}) \leq \lim_{x \rightarrow \infty}(-e^{-7x} \cos x) \leq \lim_{x \rightarrow \infty}(-e^{-7x})$$

Which means that

$$0 \leq \lim_{x \rightarrow \infty}(-e^{-7x} \cos x) \leq 0$$

Which. by the Squeeze Theorem means that

$$\lim_{x \rightarrow \infty}(-e^{-7x} \cos x) = 0$$
?

Dick
Homework Helper
Okay, I can see that $$\lim_{x \rightarrow \infty} e^{-7x} = 0$$ Which means that the limit as a whole should be 0, if it's not that case that the limit DNE. In that case, can I use the Squeeze Theorem like this?
$$[-1 \leq \cos x \leq 1] \Rightarrow [-e^{-7x} \leq e^{-7x} cos x \leq e^{-7x}]$$ Meaning that $$[\lim_{x \rightarrow \infty}(-e^{-7x}) \leq \lim_{x \rightarrow \infty}(-e^{-7x} \cos x) \leq \lim_{x \rightarrow \infty}(-e^{-7x})$$

Which means that

$$0 \leq \lim_{x \rightarrow \infty}(-e^{-7x} \cos x) \leq 0$$

Which. by the Squeeze Theorem means that

$$\lim_{x \rightarrow \infty}(-e^{-7x} \cos x) = 0$$
?

Yes, you can. There's a typo or two in there, but that's ok. I know what you mean.