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Let $h_{n}(x)=\frac{nx}{1+nx^{2}}$ for each $n\in \mathbb{N}$.

I can see that $(h_{n})$ converges pointwise to the function defined by $g(x)= \frac{1}{x}$ on the positive reals.

But I'm struggling to prove that $g$ is not the uniform limit on this domain. So far, all I've done is simplify so that $|h_{n}(x) - g(x)|= |\frac{nx}{1+nx^{2}} - \frac{1}{x}|=|\frac{nx-(\frac{1}{x}+nx)}{1+nx^{2}}|=|\frac{1}{x+nx^{3}}|$.

But I can't see how this may lead to a contradiction of the following statement (which is my definition of uniform convergence):

For any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $|h_{n}(x) - g(x)|<\epsilon$ for all $n\geq N$ and $x>0$.

How do I prove the negation of this quantified sentence? I think I'll need to consider the values of $x$ approaching $0$, but I'm unsure.

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Hint: Prove that the uniform limit of bounded functions is bounded and note that $h_n$ is bounded for each $n$, but $g$ is not.