I need to check this for uniform convergence
$$ f_n(x) = \frac{nx}{n+1} , \quad (x \in \mathbb R)$$
Here's what I did so far. (edited due to a comment telling me the way)
$$ \sup_{x \in \mathbb R} | f(x) - f_n(x) | = \sup_{x \in \mathbb R} \; \left| \frac{x}{n+1} \right|$$ $$ g(x) = \left|\frac{x}{n+1} \right| = \sqrt{\left(\frac{x}{n+1}\right)^2}$$ $$ g'(x) = \frac{x(n+1-x)}{(n+1)^3} \cdot \frac{1}{\sqrt{\left(\frac{x}{n+1}\right)^2}} \; , \quad (x \neq 0)$$ $$ g'(x) = 0 \rightarrow x=n+1 \; , \quad (x \neq 0)$$ $$ \sup_{x \in \mathbb R} | f(x) - f_n(x) | = n \nless \epsilon$$
So $f_n(x)$ does not converge uniformly! Is what I did correct?
We can see easily that $f(x)=x$ is a pointwise limite of $f_n(x) = \frac{nx}{n+1}$ and we have: $$|f_n(x)-f(x)|=\frac{|x|}{n+1}$$ so the sequence $(f_n)$ is uniformly convergent in every compact of $\mathbb{R}$ to $f$ but we have not the uniform convergence in $\mathbb{R}$ since $$||f_n-f||_\infty=\sup_{x\in\mathbb{R}}|f_n(x)-f(x)|=+\infty$$