# Solving lim x->∞ e^x cos x

• Jun 10th 2013, 03:37 PM
gesoocreesto
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 :)
• Jun 10th 2013, 03:57 PM
Plato
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 $\displaystyle \lim_{n\to\infty}f(a_n)=0~?$

• Jun 10th 2013, 04:09 PM
gesoocreesto
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 :(
• Jun 10th 2013, 04:44 PM
Plato
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.
• Jun 10th 2013, 06:14 PM
gesoocreesto
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

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

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