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

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