I came across this function: $$f(x)=\lim_{n\to \infty}{({e^{\frac{1}{x}})\arctan{1\over{1+x}}}\over{x^2+e^{nx}}}$$
To get rid of the limit, we treating $x$ as a constant,when $x >0, e^{nx} \to +\infty, \ f(x)=0$.
However, what confuses me is that $x$ is not a constant at its essence. when $x\to0^+$, $e^{1 \over x} $ goes to infinity and simply conclude $e^{nx} \to+\infty$ is a bit reckless to me. Is there any explanation for that?
On
Well if $x$ is fixed, no matter how close to $0^+$ and no matter how large $e^{1/x}$ is, the function is defined as a limit when $n\to\infty$. $n$ acts here as a dummy variable, and to see why the limit is 0 just take an enormous value, like for instance $n=1/x^3$ in your case. You'll see that for every $x$ you take, there is an $n$ such that the expression inside the limit is arbitrarily small.
\begin{align*} f(1)&=\lim_{n\rightarrow\infty}\dfrac{e^{1/1}\tan^{-1}(1/(1+1))}{1^{2}+e^{n}}=0\\ f(0.5)&=\lim_{n\rightarrow\infty}\dfrac{e^{1/0.5}\tan^{-1}(1/(1+0.5))}{(0.5)^{2}+e^{(0.5)n}}=0\\ f(x)&=\lim_{n\rightarrow\infty}\dfrac{e^{1/x}\tan^{-1}(1/(1+x))}{x^{2}+e^{nx}}=0 \end{align*} for $x>0$ is then perceived in the similar way.