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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 $x\in [-1,1]$

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    @amwhy thanks for the edit, any ideas how to solve this? – 2017-02-16
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    First look at $x > 0$ and compute the limit of $\log f_{n}(x)$. BTW, the inference that $(x^{2})^{1/2}$ depends on agreeing that we mean the positive branch of the square root. – 2017-02-16
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    is the answer that has been provided correct? or is there another way of going about this – 2017-02-16

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

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    is this a limit calculation? – 2017-02-16
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    Yes. There is an arrow $\to$ which shows the convergence of the corresponding sequence (with $x$ fixed) . – 2017-02-16
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    okay so what about the part where it says $f(x)=|x|$ for $x\in [-1,1]$? – 2017-02-16
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    I added something. Think whether the above argument is OK. – 2017-02-16
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    looks good to me I can see that it converges to |x| – 2017-02-16
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    so I don't need to do anything with the values [-1,1]? – 2017-02-16