Let $f,g:\mathbb{R}\rightarrow \mathbb{R}$. Their composition is the function $g\circ f:\mathbb{R}\rightarrow \mathbb{R}$ is defined by

$(g\circ f)(x)=g(f(x))$

Let $f$ and $g$ be functions $\mathbb{R}\rightarrow \mathbb{R}$ with $f$ continuous at some point $a\in\mathbb{R}$, and $g$ continuous at the point $f(a)$. Then $g\circ f$ is continuous at $a$.

Proof: For each $\varepsilon>0$, we must find a $\delta>0$ such that

$|x-a|<\delta\Rightarrow |(g\circ f)(x)-(g\circ f)(a)|<\varepsilon$

Let $\varepsilon>0$, since $g$ is continuous at $f(a)$, $\exists\,\delta_g>0$:

$|t-f(a)|<\delta_g\Rightarrow |g(t)-g(f(a))|<\varepsilon$

But also $f$ is continuous at $a$, so (we can get $f(x)$ $\delta_g$-close to $f(a)$), $\exists\,\delta>0$ such that

$|x-a|<\delta\Rightarrow |f(x)-f(a)|<\delta_g$

So therefore,

$|x-a|<\delta\Rightarrow |f(x)-f(a)|<\delta_g\Rightarrow |g(f(x))-g(f(a))|<\varepsilon$

$\Box$