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If we let $f_n(x)=\frac{x^2+nx}{n}$ for $x\in R$. I am trying to show that,

1) $f_n$ converges pointwise to $f(x)=x$ I believe I have done this part correctly by taking the limit and getting the following,

$$f_n(x)\frac{x^2+nx}{n}= x \left(\frac{x+n}{n} \right)=x \left(\frac{\frac{x}{n}+\frac{n}{n}}{\frac{n}{n}}\right )=x \rightarrow x$$

Just wondering if this logic is correct as I am trying to figure out how to do problems with pointwise convergence

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    Yeah, that's correct reasoning for the pointwise convergence. You might want to handle $x=0$ as a separate case but don't really need to here. – 2017-02-16
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    sorry, I was actually just editing another question, and misplaced the title – 2017-02-16

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How did you get that

$$x\frac{\frac xn+\frac nn}{\frac nn}=x\left(\frac xn+1\right)=x\;??\;\;\text{The correct symbol is certainly}\;\;\neq...$$

Anyway, the proof of that limit would be easier and clearer (and correct), I believe, remarking only that

$$\frac{x^2+nx}n=\frac{x^2}n+x\xrightarrow[n\to\infty]{}0+x=x$$

Now:

$$\sup_{x\in\Bbb R}\left|f_n(x)-x\right|=\sup_{x\in\Bbb R}\frac{x^2}n\rlap{\;\;\;\;/\;}\xrightarrow[n\to\infty]{}0\;,\;$$

and I don't get uniform convergence on the real line since the rightmost expression above isn't bounded... Check this

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    (+1) for the well-written and instructive answer - with a bonus part on UC. – 2017-02-16
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    @DonAntonio thanks for the submission that logic for the limit makes sense. I have submitted another question and have made an attempt at showing that it is not uniformly convergent but not quite sure how to finalize it – 2017-02-16