Continuity of a composition of continuous functions

Question

Suppose $f: \mathbb{R} \to \mathbb{R} $ is continuous at $x = 1 $ and $g: \mathbb{R} \to \mathbb{R} $ continuous at $y = f(1) $. Then $g \circ f $ is continuous at $x = 1 $

Attempt:

Let $\epsilon > 0$ be given. We can find $\delta', \delta'' $such that if $|x - 1| < \delta' $, then $|f(x) - f(1) | < \epsilon $ and we also have that $|f(z) - f(1) | < \delta'' $ implies $| g( f(z) ) - g( f(1)) | < \epsilon $

Isn't this just the proof? It is obvious that feel confused trying to write the proof nicely. IS this a correct proof ?

Answer 1

You are on the good track!

Let $\epsilon_1,\epsilon_2>0$ be arbitrary, by definition of continuity we have

• $f$ continuous at $1$ implies that there exists $\delta_1 >0$ such that $$|x-1|<\delta_1 \implies |f(x)-f(1)|<\epsilon_1$$

• $g$ continuous at $u$ implies that there exists $\delta_2>0$ such that $$|z-u|<\delta_2 \implies |g(z)-g(u)|<\epsilon_2$$.

Now, choose $\epsilon_1=\delta_2$ and merge the definitions.