1. ## 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.

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

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

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

Put:

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

then:

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

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.

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

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

RonL