# Limit of arctan(e^x)?

• Nidhogg
In summary, the limit of e^-7x cos x as x approaches infinity can be found using the Squeeze Theorem. By showing that the product of e^-7x and cos x is bounded between -e^-7x and e^-7x, and using the fact that the limit of e^-7x is 0, we can conclude that the limit of e^-7x cos x is also 0. Therefore, the Squeeze Theorem allows us to find the limit of this function even though one of the individual limits does not exist.

#### Nidhogg

(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.

Nidhogg said:
(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!

Dick said:
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

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

Nidhogg said:
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.