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If we let $f_n(x)=\frac{x^2+nx}{n}$ for $x\in \mathbb{R}$. I have proved that $f_n$ converges pointwise to $f(x)=x$

Now I am trying to prove that $f_n$ does not converge uniformly on $\mathbb{R}$

$$|f_n(x)-f(x) | = \sup\left|\frac{x^2+nx}{n}-x\right| = \sup \left|\frac {x^2}{n}\right|\rightarrow0$$

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    The problem in your last step is that $$\sup_{x\in\Bbb R}\frac{x^2}n$$ is **not** finite so its limit when $\;n\to\infty\;$ isn't zero, as needed to have *uniform* convergence. – 2017-02-16
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    so this solution and logic is incorrect? – 2017-02-16
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    @DonAntonio how is limit not 0? – 2017-02-16
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    Before taking the limit you **must have** the supremum, which is.... There you go!! – 2017-02-16
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    @DonAntonio looking at the answer that has been provided, does my comment to this answer make sense? – 2017-02-16

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Let $\epsilon=1$. Then, for any $n$, we select $x=\sqrt{n}$ and find that

$$\left|\frac{x^2+nx}{n}-x\right|=\left|\frac{x^2}{n}\right|\ge 1$$

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    does this prove along with my answer provided that it is not uniformly convergent? in my answer I have said that the limit must go to 0. but this answer you have provided shows a contradiction so its not uniformly convergent? – 2017-02-16
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    @user284 In your question you put some strange things, like "...using the fact that $\;x^2\rightarrow\sqrt n\;$ " (??), and then some $\;x_n\;$ ...(?). I think the above answer is what you need in the spirit of what you wanted. +1 – 2017-02-16
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    @user284 This is precisely the negation of uniform convergence for $x\in\mathbb{R}$. – 2017-02-16