If I have a sequence of functions $f_n[0,2] \rightarrow \mathbb{R}$ where $f_n(x) = \frac{x^n}{2^n+n}$.
If I attempt to find the pointwise limit, I work out that by taking $x \in [0,2]$:
We can separate the values of x, first let $0 \leq x \leq 1$, then $$ \lim_{n \rightarrow \infty} f_n(x) = \lim_{n \rightarrow \infty} \frac{x^n}{2^n+n} = \frac{\lim_{n \rightarrow \infty} x^n }{\lim_{n \rightarrow \infty} 2^n +n} =0 $$
Further we can take the scenario where $x=1$:
$$ \lim_{n \rightarrow \infty} f_n(x) = \lim_{n \rightarrow \infty} \frac{1^n}{2^n+n} = \frac{\lim_{n \rightarrow \infty} 1^n }{\lim_{n \rightarrow \infty} 2^n +n} =0 $$
Where $1 < x < 2 $, we have: $$ \lim_{n \rightarrow \infty} f_n(x) = \lim_{n \rightarrow \infty} \frac{x^n}{2^n+n} = \lim_{n \rightarrow \infty} \frac{\frac{x^n}{2^n}}{\frac{2^n+n}{2^n}} = \lim_{n \rightarrow \infty} \frac{(\frac{x}{2})^n}{1+\frac{n}{2^n}} $$
Whereby applying L'Hopitals rule we can see that $\lim_{n \rightarrow \infty} (\frac{n}{2^n})=0$ So for $1 < x < 2 $, $\lim_{n \rightarrow \infty} f_n(x)=0$
Then finally we consider the scenario where $x=2$: $$\lim_{n \rightarrow \infty} f_n(x)= \lim_{n \rightarrow \infty} f\frac{2^n}{2^n+n} = \lim_{n \rightarrow \infty} \frac{\frac{2^n}{2^n}}{\frac{2^n+n}{2^n}}= \frac{\lim_{n \rightarrow \infty} 1 }{\lim_{n \rightarrow \infty} 1+ \lim_{n \rightarrow \infty}\frac{n}{2^n}} =1$$
Thus, our final result suggests that the pointwise limit is:
$\lim_{n \rightarrow \infty} f_n(x) = \begin{cases}0,\ \ 0 \leq x<2 \\ 1,\ x=2 \end{cases} $
Despite me getting an answer, I feel like the work is inaccurate. Can anyone assist me? Thank you