$f_n:(0,\infty) \to\mathbb{R}, f_n(x)=\frac{1}{1+nx}$ uniform convergence

Question

If $(f_n)_{n\in \mathbb{N}}$ is pointwise convergent: Is the limit function continuous? Is it uniform convergent?

1. $f_n:\mathbb{R} \to\mathbb{R}, f_n(x)=xsin(nx)$

2. $f_n:(0,\infty) \to\mathbb{R}, f_n(x)=\frac{1}{1+nx}$

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1. diverges

2. $\lim_{n\to\infty}(\frac{1}{1+nx})=0$, so this is pointwise convergent and also continuous. I don't know how to prove that it's uniform convergent. How do I continue with $0-\frac{1}{1+nx}<\epsilon$?

Answer 1

It is not uniformly convergent, since it converges pointwise to $0$, but $(\forall n\in\Bbb N):f_n\left(\frac1n\right)=\frac12$.