limit and uniform convergence of sequence of function

Question

I have a sequence of function $g_n$:

$$ g_n(x) = xe^{-nx}, \qquad \text{for } x \in [0,\infty ) $$

I need to find the limit,

and determine the uniform convergnece of the sequence.

I think the limit is 0, but having trouble determining whether the convergence is uniform.

Answer 1

A reasonable conjecture for the (pointwise) limit is $g(x)=0$. So let's see if $g_n\to g$ uniformly on $[0,\infty)$. But this means $$ \max_{x\in[0,\infty)}|g_n(x)-g(x)|\to 0\text{ as }n\to\infty. $$ Just a little calculus reveals that the maximum occurs at $x={1\over n}$ so $$ \max_{x\in[0,\infty)}|g_n(x)-g(x)|=g_n(1/n)={1\over e\,n}\to 0\text{ as }n\to\infty. $$ Thus, $g_n(x)\to g(x)=0$ uniformly on $[0,\infty)$.

Answer 2

$g_n$ tends to a zero function and it converges uniformly. The first is obvious because the exponential term is less than or equal to $1$ for every $x$ and multiplying by $x$ when $x=0$ gives $0$ at $x=0$. For $x>0$ the exponential term tends to $0$ so $g_n\rightarrow 0$ pointwise.

It converges uniformyl because when you take the derivative of the function make it equal to zero. you will get $x=1/n$. When you insert it back you will get $a_n=1/(en)$ and since $a_n\rightarrow 0$ as $n\rightarrow \infty$, the convergence is uniform.