Prove that $h_n(x) = n^2 (e^{x/n} -1 - x/n)$ does not converge uniformly on $\mathbb{R}$

Question

$h_n(x) = n^2 (e^{x/n} -1 - x/n)$

So I know that the pointwise limit is $x^2/x.$ I found it by using L'hopital's rule (is there a way to show it without using l'hopital's rule?).

I know to show that the convergence is not uniform I need to:

• Find $\epsilon >0$ st $\forall N \in \mathbb{N}, \exists \geq N, x \in \mathbb{R}$ such that $|h_n(x)-x^2/n| = |n^2 (e^{x/n} -1 - x/n) - x^2/2| \geq \epsilon$.

Is the negation above correct?

From there I said let $\epsilon =1/10$ then $\forall N \in \mathbb{N},$ let $n=N$ and $x=0$ then $|n^2 (e^{x/n} -1 - x/n) - x^2/2| = |n^2 (1-1-0) - 1/2| = |-1/2| = |1/2| \geq 1/10.$

edit: that dosent work.

Answer 1

Let us notice (with the aid of Taylor series) that $h_n(x) \to \frac{x^2}2$ for any fixed $x$. So, $h_n(x)$ converges uniformly iff $\sup_{x} |h_n(x) - \frac{x^2}2| \to 0$, $n \to \infty$.

But if $x = n$ then $|h_n(x) - \frac{x^2}2| = |n^2(e - 1 -1 ) - \frac{n^2}2|$ doesn't converge to $0$. Hence, $h_n(x)$ doesn't converges uniformly.

Answer 2

Since$$n^2(e^{x/n}-1-x/n)=x^2/2+x^3/(6n)+o(x^3/n),$$the error term $\sim x^3/(6n)$ exceeds $\epsilon$ for $x\gtrsim\sqrt[3]{6n\epsilon}$.