# Finding the root, domain, and limit to infinity of $f(x) = xe^{-x}$

I had a calculus exam yesterday and the teacher asked the following question:

Find the root and domain of: $$f(x) = xe^{-x}$$ Also, find $\lim_{x\to\infty}f(x)$.

But from what I have researched, I didn't find that for $e^{ x}$ have roots, if someone know how to find the roots, thank you very much.

• That limit is a number, not a function, so I don't understand why a question is asking for roots and domain as if it were a function.
– anon
Jun 27 '13 at 1:05
• If you want to determine the root of the function, then don't indicate limits. Limits are used to determine the value at the specific x-coordinate. Jun 27 '13 at 1:11
• @NasuSama: don't edit the question to remove the question about roots or limits without confirmation/clarification from the OP. Jun 27 '13 at 1:14

I'm going to assume you mean your question to be the following:

Find the root and domain of: $$f(x) = xe^{-x}$$ Also, find $\lim_{x\to\infty}f(x)$.

First, we find the root by setting $f(x) = 0$: $$xe^{-x} = 0$$ $$\frac{xe^{-x}}{e^{-x}} = \frac{0}{e^{-x}}$$ $$x = 0$$ This is the only root. (We are justified dividing by $e^{-x}$ because $e^{-x} \ne 0$ for all $x\in\mathbb{R}$.)

The domain is $\mathbb{R}$. I don't know exactly how you'd want to "prove" this, as it's kinda self-evident (there's no way to have a not-defined value...)

To find the limit: $$\lim_{x\to\infty}xe^{-x} = \lim_{x\to\infty}\frac{x}{e^x}$$ Here, we can "eyeball" this and note that $e^x$ beats the (insert favorite word (that fits) here) out of $x$, in terms of how fast it grows. So, as $x$ gets bigger and bigger, the denominator will get bigger much faster than the numerator. Thus the limit is $0$:

$$\lim_{x\to\infty}\frac{x}{e^x} = 0$$

• Very Thanks! You put just what the teacher asked in the exam, is that I thought that joining would give the same, but not. Actually the question was $Given the function: f(x)= xe^{-x} determine, a)The domain and the roots of f, \lim{x\to+∞}f(x))$ Jun 27 '13 at 1:40