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How do you prove the following?

Suppose $y=f(x)$ such that $\lim_{x\to a}f(x) = b$

If

1) $g$ is continuous at $b$

or

2) $f(x) \neq b$ for all $x$ and $\lim_{y\to b}g(y)$ exist then $$ \lim_{x\to a}g(f(x)) = \lim_{y\to b}g(y) $$

I got the first condition its pretty straightforward:

Since $|f(x)-b|< \varepsilon$ for any $\varepsilon$, $|x-a| < \delta$ can be satisfied and $g$ is continuous $\Rightarrow$ there exist $\varepsilon$ such that $|g(f(x)-g(b)|< \varepsilon$2 is satisfied. $\varepsilon$ can always be chosen, and thus $\delta$, which gives our result for the first condition.

Im having problems with the second condition though.

Jan Lynn
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  • Zhanfeng, I formatted your question using a web-variant of LaTeX. Please make sure I didn't change the meaning of your question. – Eckhard Jan 14 '13 at 08:35

1 Answers1

1

Let $c = \lim_{y\to b} g(y)$. Let $\epsilon > 0$. Since the limit exists, there is some $\delta$ such that $0<|y-b|<\delta \Rightarrow |g(y)-c| < \epsilon.$

Since $\lim_{x\to a} f(x) = b$, there is some $\delta_2$ such that

$$0<|x-a|<\delta_2 \Rightarrow 0 < |f(x)-b| < \delta \Rightarrow |g(f(x))-c| < \epsilon,$$ and therefore $\lim_{x\to a} g(f(x)) = c.$

The inequality $0 < |f(x)-b|$ follows from $f(x) \neq b$.

user7530
  • 49,280
  • Could you clarify why the inequality 0<|f(x)−b| is necessary for the second condition, but not the first? I.e what is the difference between proofs for the first and second conditions? – Jan Lynn Jan 14 '13 at 11:10
  • $|g(y)-c|<\epsilon$ only if $0<|y-b|<\delta$. The definition of the limit of $g(y)$ says nothing if $0=|y-b|$, and indeed perhaps $g(b) \neq c$. – user7530 Jan 14 '13 at 15:56
  • Why is this not required for the first condition? – Jan Lynn Jan 16 '13 at 14:46
  • If $g$ is continuous, $\lim_{y\to b} g(y) = g(b)$ (by definition of continuity, really) so there's no worry about the case $y=b$. – user7530 Jan 16 '13 at 17:15