Is this sequence convergent or divergent? I first thought it was convergent due to pointwise limits existing at every value of $x$ but now im not sure wether or not I am right.
My sequence is $f_n(x)=\dfrac{x^n}{1+x^n} $ with $x \in [0,1]$.
How do I decide whether this sequence is convergent or divergent?
On
Note that for $x \in [0, 1)$, you have $x^n \to 0$ and thus $f_n(x) \to \dfrac 0 {1+0} = 0$. When $x=1$, you have $f_n(x) = \dfrac 1 2$. Therefore, $f_n \to f$ where $f = \left\{ \begin{matrix} 0, & x \in [0, 1), \\ \dfrac 1 2, & x = 1. \end{matrix}\right.$
On
The question says:
"I first thought it was convergent due to pointwise limits existing at every value of $x$".
That is precisely the meaning of the statement that $f_n$ converges pointwise as $n\to\infty$.
In comments, you wrote:
However doesnt the fact that the series is not continuous mean it wouldn't be convergent ??
The discontinuous nature of the pointwise limit does not change the fact that it is the pointwise limit.
It does mean that the sequence does not converge uniformly. Uniform convergence of $f_n$ to $g$ as $n\to\infty$ means $$ \lim_{n\to\infty} \sup \{ |f_n(x) - g(x)| : x\in[0,1] \} = 0. $$ That doesn't happen when $g$ is the pointwise limit. And the uniform limit, if it exists, will always be the same as the pointwise limit. Therefore, there is no function to which this sequence converges uniformly.
You may observe that $$ 0\leq f_n(x)=\frac{x^n}{1+x^n}\leq x^n, \quad x \in [0,1], $$ giving, for $x \in [0,1]$, $$f_n(x) \to f(x)=\begin{cases} 0, & \text{if $\,0\leq x<1,$} \\[2ex] 1/2, & \text{if $\,x=1.$} \end{cases}$$