# Limit of e

• October 24th 2006, 02:41 AM
aussiekid90
Limit of e
Let f be the function given by f(x)=2xe^(2x)

a) Find Lim as x reaches infinity and negative infinity (i.e. Lim )
x->infinity/-infinity

b) What is the range of f?

Thanks a lot for the help, e always confuses me and i have no idea how i should do this problem.
• October 24th 2006, 04:45 AM
CaptainBlack
Quote:

Originally Posted by aussiekid90
Let f be the function given by f(x)=2xe^(2x)

a) Find Lim as x reaches infinity and negative infinity (i.e. Lim )
x->infinity/-infinity

As both $x$ and $e^{2x}$ go to infinity as $x$ goes to $+ \infty$ clearly the

$\lim_{x \to \infty} 2x e^{2x}=\infty$.

Now as $x \to -\infty,\ x \to -\infty$ and $e^{2x} \to 0$, so we gave an indeterminent form, so we apply L'Hopitals rule

Put:

$
f(x)=\frac{2x}{e^{-2x}}
$

then:

$
\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2(-2)}{e^{-2x}}=0
$

Quote:

b) What is the range of f?

Thanks a lot for the help, e always confuses me and i have no idea how i should do this problem.
• October 24th 2006, 04:51 AM
CaptainBlack
Quote:

Originally Posted by aussiekid90
b) What is the range of f?

Thanks a lot for the help, e always confuses me and i have no idea how i should do this problem.

We know that $f(x)$ goes to $+\infty$ as $x$ becomes large , and goes to $0$
as $x$ goes to $-\infty$. Also differentiating and setting the derivative to zero
shows that $f(x)$ has a minimum of $-e^{-1}$ at $x=-1/2$.

So the range of f is $[-e^{-1}, +\infty)$.

RonL