Pointwise convergence to absolute value function

Question

I am trying to prove that,

$$f_n(x) = x^{\large {1+\frac{1}{2n-1}}} $$

converges point-wise to $f(x)=|x|$ for $x\in [-1,1]$

My thinking was to prove that it converges point-wise by taking the limit of $f_n(x)$.

$$\lim_{n\to \infty}f_n(x) = \lim _{n\to\infty}x^{\large{1+\frac{1}{2n-1}}} = |x| \lim _{n\to \infty} 1^{\large{1+\frac{\frac{1}{n}}{\frac{2n}{n}-\frac{1}{n}}}} = |x| \lim _{n\to \infty} 1 = |x|$$

But from here how do I prove that it converges point-wise to $f(x)=|x|$ for $x\in [-1,1]$

Answer 1

$$ f_n(x)=x^{1+1/(2n-1)}=x^{2n/(2n-1)}=\sqrt[2n-1]{x^{2n}} =(x^2)^{n/(2n-1)} \to (x^2)^{1/2}=|x| $$