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  1. $\lim \limits_{x \to a}{f(x)} = f(a)$ and $\lim \limits_{y \to b}{g(y)} = a$, then $\lim \limits_{y \to b}{f(g(y))} = f(a)$
  2. $\lim \limits_{x \to a}{f(x)} = f(a)$ and $\lim \limits_{y \to b}{g(y)} = a$, then $f(g(b) = f(a)$
  3. $\lim \limits_{x \to g(b)}{f(x)} = f(g(b) )$ and $\lim \limits_{y \to b}{g(y)} = a$, then $\lim \limits_{y \to b}{f(g(y))} = f(a)$
  4. $\lim \limits_{x \to g(b)}{f(x)} = f(g(b) )$ and $\lim \limits_{y \to b}{g(y)} = a$, then $f(g(b)) = f(a)$
  5. $\lim \limits_{x \to g(b)}{f(x)} = f(g(b) )$ 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?

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    That isn't enough as a proof as it is assuming the result you are trying to prove. – 2017-01-12
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    See if this works for $g(y) = \begin{cases}a, & y\neq b \\ a+1, & y = b\end{cases}$ – 2017-01-12
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    Rather than asking five problems at once, you need to start with your misunderstanding of the first problem. $f$ is not a constant that can be "factored out" of the limit as $y\to b$. – 2017-01-12
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    Well, i thought i could use $\lim_{n\rightarrow \infty} g(f(x_n))=g(\lim_{n\rightarrow \infty} f(x_n)) = g(f(\lim_{n\rightarrow \infty} x_n))$ Because i thought f(x) is continuous at a – 2017-01-12
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    Can u guys give me some more help? I'm legit going crazy, even thought this task is probably not that hard. – 2017-01-12
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    @TimBender So... you did get "some more help." Was that enough? – 2017-01-12

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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.