Consider the sequence of functions $\{f_n \}$ on $\Bbb R$ defined by $$f_n(x) = n \log \left (1 + \frac {x^2} {n} \right ),\ \ x \in \Bbb R$$ for all $n \geq 1.$ Show that the sequence of functions $\{f_n \}$ is not uniformly convergent on $\Bbb R.$
My attempt $:$ First I observe that the sequence of functions $\{f_n \}$ converges pointwise to the everywhere continuous function $f$ defined by $$f(x) = x^2,\ \ x \in \Bbb R.$$
If the sequence of functions $\{f_n \}$ converges uniformly to the continuous limit function $f$ on whole of $\Bbb R$ then in particular the sequence of functions $\{f_n \}$ converges uniformly to the continuous limit function $f$ on $[0,1].$
But then we have $$\lim\limits_{n \rightarrow \infty} \int_{0}^{1} f_n(x)\ dx = \int_{0}^{1} f(x)\ dx.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
Now what I found is that $$\int_{0}^{1} f_n(x)\ dx = n \log \left (1 + \frac 1 n \right ) - 2n + 2n^{\frac 3 2} \arctan \left (\frac {1} {\sqrt n} \right ).$$ So we have $$\lim\limits_{n \rightarrow \infty} \int_{0}^{1} f_n(x)\ dx = - \infty.$$ where as $$\int_{0}^{1} f(x)\ dx = \frac 1 3.$$ which is a contradiction to $(1).$ Hence our assumption is false. So the sequence of functions $\{f_n \}$ is not uniformly convergent on $\Bbb R,$ as required.
EDIT $:$ I found the wrong limit. By L'Hospital the limit $$\lim\limits_{n \rightarrow \infty} n \left ( \log \left (1 + \frac 1 n \right ) - 2 \right ) = 1.$$
Now what will be $$\lim\limits_{n \rightarrow \infty} 2n^{\frac 3 2} \arctan \left (\frac {1} {\sqrt n} \right )\ ?$$
By L'Hospital I got infinity. Isn't it the case? But then the overall limit becomes $+\infty.$