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$$f_n(x)=(1+x^n)^{\frac{1}{n}}, x\in[0,2]$$

Show that this sequence is uniformly convergent to a function which's not differentiable at 1.

While trying to show uniform convergence:

When $0\le x \le 1: \lim_{n \rightarrow \infty} f_n(x)=1$.

When $x\gt 1: \lim_{n \rightarrow \infty} f_n(x)=x$.

So, How is $f_n$ uniformly convergent when its pointwise limit is not continuous?

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    $\begingroup$ But the pointwise limit is continuous. $\lim\limits_{x \to 1^+} x = 1$. $\endgroup$ – Daniel Fischer Apr 23 '17 at 19:22
  • $\begingroup$ The limit is continuous. but not differentiable at x=1. $\endgroup$ – DanielWainfleet Apr 23 '17 at 22:58

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