Prove that the function sequence $f_n(x)=n^2\left(\mathrm{e}^{\frac{1}{n x}}-1\right) \sin \frac{1}{n x}$ convergent uniformly on $[a,+\infty)(a>0)$ .

Prove that the function sequence $f_n(x)=n^2\left(\mathrm{e}^{\frac{1}{n x}}-1\right) \sin \frac{1}{n x}(n=1,2, \cdots)$ convergent uniformly on $[a,+\infty)(a>0)$ .

Proof1. For every $x \in[a,+\infty)$, we have$$\lim _{n \rightarrow \infty} f_n(x)=\lim _{n \rightarrow \infty} \frac{\sin \frac{1}{n x}}{\frac{1}{n x}} \cdot \frac{\mathrm{e}^{\frac{1}{n x}}-1}{\frac{1}{n x}} \cdot \frac{1}{x^2}=\frac{1}{x^2} .$$

Notice that while $t \rightarrow 0$ , we have$$\mathrm{e}^t-1 \sim t+o(t), \quad \sin t \sim t+o(t),$$

so when $x \in[a,+\infty)$ and $n \rightarrow \infty$ , we have$$\begin{aligned}& \left|n^2\left(\mathrm{e}^{\frac{1}{n x}}-1\right) \sin \frac{1}{n x}-\frac{1}{x^2}\right|=n^2\left|\left(\mathrm{e}^{\frac{1}{n x}}-1\right) \sin \frac{1}{n x}-\frac{1}{n^2 x^2}\right| \\& \quad=n^2\left|\left[\frac{1}{n x}+o\left(\frac{1}{n x}\right)\right]\left[\frac{1}{n x}+o\left(\frac{1}{n x}\right)\right]-\frac{1}{n^2 x^2}\right| \\& \quad=n^2 o\left(\frac{1}{n^2 x^2}\right) .\end{aligned}$$

So we have$$\sup _{x \in[a,+\infty)}\left|n^2\left(\mathrm{e}^{\frac{1}{n x}}-1\right) \sin \frac{1}{n x}-\frac{1}{x^2}\right| \leqslant n^2 o\left(\frac{1}{n^2 a^2}\right) \rightarrow 0 \quad(n \rightarrow \infty) .$$

This proves $f_n(x) \rightrightarrows \frac{1}{x^2}(x \in[a,+\infty))$.

Proof2:

(a) $f_n(x) \stackrel{x \in(0,+\infty)}{\nrightarrow} f(x)$$\sup _{x \in(0,+\infty)}\left|f_n(x)-f(x)\right| \geqslant n^2|(e-1) \sin 1|\nrightarrow 0$

(b) $f_n(x) \stackrel{x \in[\delta,+\infty)}{\rightrightarrows} f(x)$

Note that $\left|0\left(\frac{1}{n x}\right)\right|=\left|0\left(\frac{1}{n \delta}\right)\right|$$$\begin{aligned}f_n(x)-f(x) & =\frac{1}{x^2}\left[\frac{e^{-n x}-1}{1 / n x} \cdot \frac{\sin \frac{1}{n x}}{1 / n x}-1\right] \\& =\frac{1}{x^2}\left[\frac{1}{t^2}\left(t+\frac{e^{\theta t}}{2 !} t^2\right)\left(t-\frac{\cos \hat{\theta} t}{3 !} t^3\right)-1\right]\left(t=\frac{1}{n x}\right) \\& =\frac{1}{x^2}\left[\frac{e^{\theta t}}{2 !} t-\frac{\cos \hat{\theta} t}{3 !} t^2+\frac{e^{\theta t}}{2 !} \frac{\cos \hat{\theta} t}{3 !} t^3\right]\end{aligned}$$

When $n$ is large enough, $\exists C \forall x \in[\delta,+\infty)\left|f_n(x)-f(x)\right| \leqslant \frac{1}{\delta^2} \cdot C \cdot \frac{1}{n \delta}\xrightarrow[n \rightarrow+\infty]{} 0$

Which of these two proof ideas is correct?