Given the functional sequence $f_n(x) = \frac{x^3}{n^2} +\sin{x}$ show that $f_n(x)$ converges pointwise to a function $f$ in $\mathbb{R}$. Next, show that $f_n(x)$ converges uniformly on the closed interval $[-10, 10]$.
My work and thoughts:
Since I’m new to this subject my first thought was to open Mathematica to get a visual representation of the above functional sequence. Looking at the graph below it is clear that the functional sequence $f_n(x)$ converges pointwise to the constant function $f(x) = 0$. Indeed, for a fixed value of x, the points $f_n(x)$ converges to zero (as $n \rightarrow \infty$).
In mathematical terms:
For $x = 0$ we have $f_n(0) = 0$ for all $n \in \mathbb{N}$.
For $x \neq 0$ how do I prove that the $\lim\limits_{n\to\infty} f_n(x) = 0$? The sinus function is bounded ($\left| \sin{x} \right| \leq 1$) so it’s negligible but I don’t know how to take care of the $\frac{x^3}{n^2}$ factor when taking the limit.
Concerning the uniform convergence I’m a little confused by the graph.
By the definition of uniform convergence the functional sequence $f_n$ is uniformly convergent to a limit function $f$ if: $$\forall{\varepsilon} > 0 \; \exists N \in \mathbb{N} \; \forall{x} \in [-10, 10] \; \forall{n} \in \mathbb{N} : n > N \implies \left| f_n(x) – f(x)\right| < \varepsilon.$$
Looking at the graph, the value of $n$ required for the inequality to hold clearly depends on $x$ and so $f_n(x)$ cannot be uniformly convergent on $[-10, 10]$.
What did I miss here and how do I show that $f_n$ converges uniformly to $f$?
