1

$$\lim_{x \to c} f(x) = L \iff \lim_{h \to 0} f(c+h) = L$$

My prove for ($\implies$) was: Let $g(h) = f(c+h)$, then we are required to show that $\lim_{h \to 0} g(h) = L$. I went ahead and tried to prove $\lim g(h) = L$ using epsilon-delta but don't know what I am supposed to do with $|g(h) - L|<e$. Any help would be great!

A. Wal
  • 23
  • See this. – edm Jul 03 '17 at 05:56
  • 1
    How did you get $|g(c)-L|$ in the first place? It would be helpful to see what you have tried so far. To me it seems like I cannot fail if I just apply the $\epsilon$-$\delta$ definition. That of course doesn't mean that other people cannot fail, but it's not that easy to predict where other people will fail... – skyking Jul 03 '17 at 06:38
  • Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jul 03 '17 at 11:02

1 Answers1

1

If $\lim_{x\rightarrow c}f(x)=L$, this implies that for every $\epsilon>0$ there is a $\delta>0$ such that $$|f(x)-L|<\epsilon\qquad \text{whenever}\qquad0<|x-c|<\delta.$$ Now putting $x-c=h$, the above relation reduces to $$|f(c+h)-L|<\epsilon\qquad \text{whenever}\qquad0<|h|<\delta.$$ This means $\lim_{h\rightarrow 0}f(c+h)=L$.

Retracing the steps backward, we get the other way implication.

mark haokip
  • 573