How to prove this using the epsilon delta definition?

Question

$\lim_{x \to a} f(x)=\lim_{h \to 0} f(h+a)$

The above seemingly obvious result is what I want to prove.

My idea was to "set" $x$ to be equal to $h+a$ and then prove using the epsilon delta definition that $\lim_{h \to 0} h+a=a$ or that $\lim_{h \to 0} x=a$. This would (i think) mean that $\lim_{x \to a} f(x)=a$ since $x=h+a$ and ${h \to 0}$ implies that ${x \to a}$.

I didn't like it because we somehow had to "set" $x$ equal to $h+a$ in the beginning. I wonder if there is a better way to prove it.

Answer 1

Let $\lim_{x\rightarrow a}f (x)=c $ (assuming limit exists). Let $\epsilon >0$. Then there exists a $\delta >0$ so that $|x-a|<\delta \implies |f (x)-c|<\epsilon$. Then $|h-0|=|(h+a)-a|<\delta \implies |f (h+a)-c|<\epsilon $.

So $\lim_{h\rightarrow 0}f (h+a)=c $.

Most do same thing for when limit doesn't exist.

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I see by your comment you didn't understand this

Definition: $\lim_{w\rightarrow a} f(w) = c$ means the for any value $\epsilon 0$ there exist a $\delta >0$ so that the following is true: If $|w - a| < \delta$ or in other words if $w \in (a - \delta, a + \delta)$ then $|f(w) - c| < \epsilon$ or in other words $f(w) \in (c-\epsilon, c + \epsilon)$.

Now $w \in (a-\delta, a+ \delta) \iff w-a \in (0 -\delta, 0+ \delta)$.

So $\lim_{x\rightarrow a}f(x) = c$ means for any $\epsilon > 0$ there exists a $\delta$ so that $a-x \in (0-\delta, 0+\delta) \implies f(x)=f((x-a) + a) \in (c - \epsilon, c + \epsilon)$.

And that is the definition of $\lim_{x - a \rightarrow 0}f((x-a) +a) = c$. Now we can just relabel $x - a$ as $h$ where $h = x - a$. So $\lim_{h \rightarrow 0}f(h + a) = c = \lim_{x \rightarrow a} f(x)$.

Answer 2

With $a,L\in\mathbb R, $and $f:I\to\mathbb R,\;$we have

$\lim_{x\to a}f(x)=L \implies$

$\forall \epsilon>0\;\;\exists \alpha>0\;\;\forall x\in I $

$a-\alpha<x<a+\alpha \;\;\implies |f(x)-L|<\epsilon$

$\implies$

$\forall \epsilon>0 \;\;\exists \alpha>0; \forall h>0\;\;|h|<\alpha$

$\implies a-\alpha<a+h<a+\alpha $

$\implies |f(a+h)-L|<\epsilon $

$\implies \lim_{h\to 0}f(a+h)=L$

$\implies$

$\forall \epsilon>0\;\exists \alpha>0\;\forall x\in I a-\alpha<x<a+\alpha \implies |h|=|x-a|<\alpha \implies |f(a+x-a)-L|=|f(x)-L|<\epsilon$

$\implies \lim_{x\to a}f(x)=L.$