Can someone explain to me why ${f_n}=x^{\frac{1}{2n-1}}$ defined on [-1,1] converges pointwise to $1$ if $x \in (0,1)$, to $-1$ if $x \in (-1,0)$, and to $0$ if $x=0$. This is the answer I got but I think it should converge to $1$ if $x \in [-1,1]-\{0\}$. Because there is an $N$ for which the exponent gets extremely small so it approaches $0$ (so anything raised to the $0$ power is $1$).
An intuitive explanation or a proof would be really appreciated.
So you're only confused about the convergence on $(-1,0)$, yes? The answer says pointwise convergence to $-1$ but you say $1$, yes? I will address this confusion below with both a proof, and an intuitive explanation.
Mathematically, we have convergence to $-1$ via the following. Let $x\in(-1,0)$, then $$\lim_{n\to\infty}x^{1/(2n-1)}=-\lim_{n\to\infty}|x|^{1/(2n-1)}=-(1)=-1$$ since the exponent $\frac{1}{2n-1}$ is always an odd root, so we can extract the $-1$ and just look at $|x|\in(0,1)$. Notice that we need to consider the limit with $|x|$ so that we can use continuity of $y\mapsto b^y$ for a fixed positive base $b$; if $b<0$ then $y\mapsto b^y$ is not continuous.
Intuitively, we can think about the following. Take some number $x\in(-1,0)$. Then when we take odd roots of $x$ we don't change its sign, because $(-1)^{1/(2n-1)}=-1$. So as $n$ grows $|x|$ is getting closer and closer to $1$, as we can see by the argument you give, however the sign of $x$ is always negative as $n$ grows, so we ultimately end up at $-1$.