investigate uniform convergence of the sequence of functions $f_{n}(x)$ = $\frac{(n+1)x + n^{2}x^{3}}{1 + n^{2}x^{2}}$. investigate uniform convergence of the sequence of functions $f_{n}(x)$ = $\frac{(n+1)x + n^{2}x^{3}}{1 + n^{2}x^{2}}$.
i am asked to investigate uniform convergence of the sequence of functions provided about. should I use Weierstrass M-Test?
I know that If $x$ $\neq$ 0, $\lim_{n \to \infty} f_{n}(x)$ = $x$ ($pointwise$ $limit$)
 A: As per the suggestions in my comments, we can solve this problem by seeing if the maximum of $|f_n(x)-f(x)|$ goes to zero as $n\to\infty$. As you have pointed out, the pointwise limit of the sequence $(f_n)$ is the identity function. Hence,
$$f_n(x)-f(x)=\frac{(n+1)x+n^2x^3}{1+n^2x^2}-x=\frac{nx+x+n^2x^3-x-n^2x^3}{1+n^2x^2}=\frac{nx}{1+n^2x^2}$$
Differentiating with respect to $x$ gives
$$\frac{d}{dx}(f_n(x)-f(x))=\frac{(1+n^2x^2)(n)-(nx)(2n^2x)}{(1+n^2x^2)^2}=\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}$$
Therefore, if the critical points of $f_n-f$ occur at $x_0$, then we have
$$\frac{n(1-n^2x_0^2)}{(1+n^2x_0^2)^2}=0\iff x_0=\pm\frac{1}{n}$$
Hence, it suffices to show $\lim_\limits{n\to\infty}\left|f_n\left(\pm\frac{1}{n}\right)-f\left(\pm\frac{1}{n}\right)\right|=0$ to show uniform convergence of $(f_n)$ to $f$. Or, if $\lim_\limits{n\to\infty}\left|f_n\left(\pm\frac{1}{n}\right)-f\left(\pm\frac{1}{n}\right)\right|\neq 0$, then we do not have uniform convergence.
Addendum: In case there is any confusion with this technique used to determine uniform convergence, let me explain. In order to see if $(f_n)$ converges uniformly to $f$, we can equivalently see if the maximum distance between $f_n$ and $f$ goes to zero as $n\to\infty$. This is precisely what the computation $\lim_\limits{n\to\infty}\sup_x |f_n(x)-f(x)|$ does: if this limit is zero then the maximum distance between $f_n$ and $f$ goes to zero (so we have uniform convergence), but if this limit is nonzero, then there is always some distance between the functions (so we cannot have uniform convergence).
