$\lim_{x \rightarrow \infty} \lim_{n \rightarrow \infty} f_{n}(x) = \lim_{n \rightarrow \infty}\lim_{x \rightarrow \infty}f_{n}(x) $ in case of uniform convergence of $f_{n}(x)$ to $f(x)$
I used this while I was given that $f_{n}(x)$ converges to $f(x)$ uniformly. Is this correct and I was also thinking to prove this!
It is correct.
Let $g=\lim_{x\rightarrow\infty}f(x)$ and $g_n=\lim_{x\rightarrow\infty}f_n(x)$. We need to show that $g = \lim_{n\rightarrow\infty}g_n$, that is $$ \forall \epsilon>0 \;\exists N \;\forall n>N: |g-g_n| < \epsilon $$ For a given $\epsilon$ let us choose $N$ such that $$ \forall n>N \;\forall x : |f(x)-f_n(x)| < \epsilon/3$$ we can do that, because $f_n$ converges to $f$ uniformly. Now, for every $n$ let us choose $x_n$ such that $$ |g-f(x_n)| < \epsilon/3 \qquad \text{and} \qquad |f_n(x_n)-g_n|<\epsilon/3$$ We can do that because for $x\rightarrow\infty$, $f$ and $f_n$ converge to $g$ and $g_n$ respectively.
Then let us note that for $n>N$: $$ |g-g_n| \le |g-f(x_n)|+|f(x_n)-f_n(x_n)|+|f_n(x_n)-g_n| < \epsilon $$