Uniform convergence with exponential function

Question

I have to study the uniform convergence of the following function for $x \geq 0$:

$$f_n(x)=\frac{ne^x+xe^{-x}}{n+x}$$ the pointwise limit is $f(x)=e^x$. For the uniform convergence i had thought to do so:

$$|f_n(x)-f(x)|=\left|\frac{xe^{-x}-xe^x}{n+x}\right|=\frac{x(e^x-e^{-x})}{n+x}$$ now, since $x \geq 0$ i can say:

$$\frac{x(e^x-e^{-x})}{n+x} \leq \frac{x(e^x-e^{-x})}{n}$$ now I don't know how to go on. Can I say that the last limit goes to $0$ as n goes to $+\infty$, so there's uniform convergence for $x \geq 0$? For $x$ too large, even the numerator goes to $+\infty$, so I don't know what to do.

I would also like to know if there's uniform convergence in $[0, b]$ with $b < + \infty$.

Answer 1

If the convegence is uniform then there exists $m$ such that $|f_n(x)-f(x)| <1$ for all $n >m$ and and for all $x \geq 0$. Take $n=m+1$. You get a contradiction by letting $x \to \infty$. Note that $\frac x {n+x} \to 1$ and $e^{x}-e^{-x} \to \infty -0=\infty$.

Answer 2

It is easy to see that $\lim_{n\to\infty}\bigl|f_n(n)-f(n)\bigr|=\infty$. Therefore, the convergence is not uniform.

On the other hand, the convergence is uniform in intervals of the type $[0,b]$. You can check that$$f(x)-f_n(x)=2\frac{\sinh(x)x}{n+x}$$and that tharefore$$f'(x)-f_n'(x)=2\frac{n\sinh(x)+x(n+x)\cosh(x)}{(n+x)^2}>0.$$So, the maximim of $f-f_n$ on $[0,b]$ is $f(b)-f_n(b)=2\frac{\sinh(b)b}{n+b}$. So, since $\lim_{n\to\infty}2\frac{\sinh(b)b}{n+b}=0$, the convergence is uniform.