- $\lim \limits_{x \to a}{f(x)} = f(a)$ and $\lim \limits_{y \to b}{g(y)} = a$, then $\lim \limits_{y \to b}{f(g(y))} = f(a)$
- $\lim \limits_{x \to a}{f(x)} = f(a)$ and $\lim \limits_{y \to b}{g(y)} = a$, then $f(g(b) = f(a)$
- $\lim \limits_{x \to g(b)}{f(x)} = f(g(b) )$ and $\lim \limits_{y \to b}{g(y)} = a$, then $\lim \limits_{y \to b}{f(g(y))} = f(a)$
- $\lim \limits_{x \to g(b)}{f(x)} = f(g(b) )$ and $\lim \limits_{y \to b}{g(y)} = a$, then $f(g(b)) = f(a)$
- $\lim \limits_{x \to g(b)}{f(x)} = f(g(b) )$ and $\lim \limits_{y \to b}{g(y)} = g(b)$, then $\lim \limits_{y \to b}{f(g(y))} = f(g(b))$
I'm somehow stuck. I thought about doing it like this: 1. $\lim \limits_{y \to b}{f(g(y))} = f\lim \limits_{y \to b}{g(y)}=f(a) $ Is that enough as a proof?
I'll do the first (with an "$\varepsilon$/$\delta$" argument) to get you started. If you are still confused for the 4 others, after thinking about them, I will add some hints.
For the first: let $\varepsilon > 0$ be arbitrary, and let $\delta>0$ be such that $0<\lvert x-a\rvert <\delta$ implies $\lvert f(x)-f(a)\rvert <\varepsilon$. This is guaranteed by the definition of the first limit; moreover, note that since $\lvert f(a)-f(a)\rvert <\varepsilon$ (!), we actually get that $0\leq \lvert x-a\rvert <\delta$ implies $\lvert f(x)-f(a)\rvert <\varepsilon$. $\textsf{(1)}$.
Similarly, let $\delta'>0$ be such that $0<\lvert y-b\rvert <\delta'$ implies $\lvert g(y)-a\rvert <\delta$. $\textsf{(2)}$
Then, for any $y$ such that $0<\lvert y-b\rvert <\delta'$, we have by $\textsf{(2)}$ that $0\leq \lvert g(y)-a\rvert <\delta$, and therefor by $\textsf{(1)}$ (applied to $x\stackrel{\rm def}{=} g(y)$) that $\lvert f(g(y))-f(a)\rvert <\varepsilon$. Since $\varepsilon>0$ was arbitrary, this shows that $$\lim_{y\to b} f(g(y)) = f(a)$$ indeed.