Check uniform convergence of a sequence

Question

Check uniform convergence:
$f_n(x) = \exp{(x+\frac{1}{\sqrt{\ln{n}}}}) + \cos({\frac{x}{(1+\frac{1}{n})^n}})$.
We know that $f_n(x) \to f(x) = e^x + \cos{\frac{x}{e}}$, so $|f_n(x) - f(x)| \leq |e^{x+\frac{1}{\sqrt{\ln{n}}}} - e^x| + |\cos({\frac{x}{(1+\frac{1}{n})^n}}) - \cos{\frac{x}{e}}|$ and I have shown that for $x = n$ as $n \to \infty$ $|e^{x+\frac{1}{\sqrt{\ln{n}}}} - e^x| \to \infty$, so it can't converge uniformly. Is it correct?

Answer 1

With the correct bounds, your idea of considering $x=n$ can be used to prove non-uniform convergence for $x \in [0,\infty)$.

For $x \in [0,\infty)$, we have

$$\begin{align}|f_n(x) - f(x)| &= \left|e^{x+\frac{1}{\sqrt{\ln{n}}}} - e^x + \cos\left({\frac{x}{(1+\frac{1}{n})^n}}\right) - \cos{\frac{x}{e}} \right| \\ &\geqslant \left|e^{x+\frac{1}{\sqrt{\ln{n}}}} - e^x\right| - \left| \cos\left({\frac{x}{(1+\frac{1}{n})^n}}\right) - \cos{\frac{x}{e}} \right|\\ &\geqslant \left|e^{x+\frac{1}{\sqrt{\ln{n}}}} - e^x\right| - 2 \\ &= e^x \left(e^{\frac{1}{\sqrt{\ln{n}}}} - 1\right)-2\\ &\geqslant \frac{e^x}{\sqrt{\ln n}}-2\end{align}$$

Thus, convergence is not uniform on $[0,\infty)$ since

$$\sup_{x \in [0,\infty)} |f_n(x) - f(x)| \geqslant \frac{e^n}{\sqrt{\ln n}}-2 \underset{n \to \infty}\longrightarrow +\infty$$