Proving pointwise convergence of sequence of functions $f_n(x) = ne^{\frac{x}{n}} - n $

Question

Letting $f_n: \mathbb{R} \to \mathbb{R}$, we define a sequence of functions where each function is: $$f_n(x) = ne^{\frac{x}{n}} - n$$ for all $n$. How do I show this sequence converges pointwise, and how do I find its limit?

I've tried graphing this sequence of functions, and it seems to approach $f(x) = x$. Using the epsilon definition to prove pointwise convergence, I tried to show that for each $\epsilon > 0$ and $x \in \mathbb{R}$, there exists $N$ such that $ |f_n(x) - x| < \epsilon$ for $n>N$, or $$|ne^{x/n} - n - x| < \epsilon $$

However, in analyzing $|ne^{x/n} - n - x|$, unless $x = 0$, I wasn't able to come up with $N$ such that $n > N$ would imply $|ne^{x/n} - n - x| < \epsilon$ for $x > 0 $ and $x<0$. I understand that for pointwise convergence, $N$ could depend on $x$ and $\epsilon$, but it seems the $n$ in the exponent of $e$ is causing me some difficulty in proving the above.

What's the approach to complete this proof? Or is there a better way to show pointwise convergence of this sequence of functions? Does it actually converge to $f(x) = x$? Is there an easy way to see the limit of this function sequence without graphing it and guessing?

Answer 1

From the Taylor expansion of $e$, we have that

$$|ne^{\frac{x}{n}} - n - x| = \Bigr|\frac{x^2}{2n} + \frac{x^3}{6n^2} + \cdots\Bigr| < \frac{x^2}{2n}\left(1 + |x| + \frac{x^2}{2} + \cdots\right) = \frac{x^2e^{|x|}}{2n} < \epsilon$$

so let $N = \left\lceil \frac{x^2e^{|x|}}{2\epsilon} \right\rceil$. Then the inequalities above follow in reverse and we get that $|f_n(x)-x| < \epsilon$.

Answer 2

For each $x\in\mathbb R$,\begin{align}\lim_{n\to\infty}ne^{\frac xn}-n&=\lim_{n\to\infty}\frac{e^{x/n}-1}{\frac1n}\\&=x\lim_{n\to\infty}\frac{e^{x/n}-1}{\frac xn}\\&=x\exp'(0)\\&=x.\end{align}