Solving lim x->∞ e^x cos x

Hi everybody :)

I'm trying to solve this limit

$\displaystyle \lim_{x\to\infty}e^x cos(x)$

Since $\displaystyle \lim_{x\to\infty}e^x = \infty$

and $\displaystyle \lim_{x\to\infty}cos(x)$ is a Real number in $\displaystyle [-1,1]$

is it correct to say that the solution is $\displaystyle \infty$?

Thank you for your help :)

Re: Solving lim x->∞ e^x cos x

Quote:

Originally Posted by

**gesoocreesto** to solve this limit

$\displaystyle \lim_{x\to\infty}e^x cos(x)$

Since $\displaystyle \lim_{x\to\infty}e^x = \infty$

and $\displaystyle \lim_{x\to\infty}cos(x)$ is a Real number in $\displaystyle [-1,1]$

is it correct to say that the solution is $\displaystyle \infty$?

Write $\displaystyle f(x)=e^x\cos(x)$. Then $\displaystyle f$ is **continuous**.

Let for $\displaystyle n\in\mathbb{Z}^+$ define $\displaystyle {a_n} = \frac{{\left( {n + 1} \right)\pi }}{2}$ is it true that $\displaystyle \lim_{n\to\infty}f(a_n)=0~?$

Now what do you think about your question now?

Re: Solving lim x->∞ e^x cos x

No, it's not true. That limit goes to infinity. But I'm sorry I don't get the point :(

Re: Solving lim x->∞ e^x cos x

Quote:

Originally Posted by

**gesoocreesto** No, it's not true. That limit goes to infinity. But I'm sorry I don't get the point :(

**It certainly is true!**

You had better check you basic knowledge to trig values.

For any odd multiple of $\displaystyle \pi/2$ the $\displaystyle \cos$ is zero.

Re: Solving lim x->∞ e^x cos x

Got it, thank you for your help :)

I ended up understanding that the solution I was searching for is based on the definition of limit superior/inferior. So

$\displaystyle \limsup\limits_{x\rightarrow\infty}e^x cos (x) = +\infty$

$\displaystyle \liminf\limits_{x\rightarrow\infty}e^x cos (x) = -\infty$