We consider the topology of the extended real line. Let $h\colon [-\infty,\infty]\to\Bbb R$ and let $(-\infty,0)$ and $(\infty,0)$ are limit point of the graph of $h$, that is, they are limit points of $\{(x,h(x))\colon x\in\Bbb R\}.$ Now, let $g\colon (a,b)\to \Bbb R$ be a homeomorphism. Notice that $\lim\limits_{x \to a^{+}}g(x)=\infty \ \ \text{or} -\infty$ and $\lim\limits_{x \to b^{-}}g(x)=\infty \ \ \text{or} -\infty.$ This is true for any homeomorphism. Now, consider $h\circ g\colon (a,b)\to \Bbb R$
Claim $$ (a,0)\ \text{and}\ (b,0)\ \ \text{are limit points of the graph}\ h\circ g$$ I think my claim would be true since homemomorphism behaves nicely with topological property.
My attempt was to consider a sequence $x_n$ in $(a,b)$ and show $(x_n, (h\circ g)(x_n))\to (a,0)$ and then in final step I have to commute between limit and $h$ to got what I need. Of course, I can not do that in general since $h$ is not a continuous function. This is why I am asking. Any help will be appreciated greatly