# Proving limit of functions involving neighborhoods

Let $A$ and $x_0$ be real numbers, and let $f(x)$ be a real-valued function defined in a deleted neighborhood of $x_0$. Use the definition of limit to prove that $\lim \limits_{x \to x_0} f(x) = A$ if and only if $\lim \limits_{x \to 0} f(x_0 + x) = A$.

so assume that $\forall \epsilon>0$ $\exists \delta=\delta(\epsilon,x_0)$ such that $|f(x)-A|<\epsilon$ for all $x$ where $0<|x-x_0|<\delta$

Therefore $x_0-\delta<x<x_0+\delta$ and $A-\epsilon<f(x)<A+\epsilon$

Can we set $\epsilon$ equal to $x_0$, since $x_0$ is a some real number?

I'm having a lot of trouble understanding this process, can anyone walk me through it?

Let $f$ be a real valued function, $x_0$ a real fixed number and $A$ a real number.

Intuitively, approaching to $x_0$ can be achieved by either taking a $x$ very close to $x_0$ (meaning $x \rightarrow x_0$) or by considering the number $x_0+x$ with $x$ a very small in absolute value (meaning $x\rightarrow 0$)

Formally, we'll first see that $$\lim_{x\rightarrow x_0} f(x)=A \Rightarrow \lim_{x\rightarrow 0} f(x_0 + x)=A$$

Let $\epsilon$ be an arbitrary fixed small positive quantity. We want to prove that there is a $\delta > 0$ (which may depend both on $x_0$ and $\epsilon$) such that $|f(x_0+x)-A|<\epsilon$ whenever $0<|x-0|=|x|<\delta$

Because $\lim_{x\rightarrow x_0} f(x)=A$, there is a $\delta' > 0$ (which depends both on $x_0$ and $\epsilon$) such that if $|x_0 - x|<\delta'$, then $|f(x) - A|<\epsilon$

Notice that $|x|=|(x+x_0) - x_0|$, so by taking $\delta =\delta'$, if $0<|x| = |(x+x_0)-x_0| < \delta$ then $|f(x+x_0) - A|<\epsilon$

Hence $\delta = \delta'$ works and we've just proved that $\lim_{x\rightarrow 0} f(x_0 + x)=A$ as we wanted.

Now we'll first see that $$\lim_{x\rightarrow 0} f(x_0 + x)=A \Rightarrow \lim_{x\rightarrow x_0} f(x)=A$$

The reasoning is pretty much the same. Again, let $\epsilon>0$ be a fixed quantity. We need to find a $\delta > 0$ such that if $0<|x -x_0| < \delta$ then $|f(x) - A|<\epsilon$

Because $\lim_{x\rightarrow 0} f(x_0 + x)=A$ there is a $\delta' > 0$ such that if $|x|<\delta'$, then $|f(x_0 + x) - A|<\epsilon$

Notice that $f(x) = f(x_0 + (x -x_0))$, so by taking $\delta=\delta'$, if $0<|x-x_0|<\delta$ then $|f(x_0 + (x-x_0))-A|=|f(x)-A|<\epsilon$

Hence $\delta =\delta'$ works and we've just proved that $\lim_{x\rightarrow x_0} f(x)=A$ as we wanted.

If $\lim_{x \to x_{0}}f(x) = A$ and if $\varepsilon > 0$, then there is some $\delta > 0$ such that $0 < |x-x_{0}| < \delta$ only if $|f(x) - A| < \varepsilon$; note that $0 < |x| < \min \{ |2x_{0}-\delta|, |2x_{0}+\delta| \} =: M$ only if $2x_{0} - \delta - x_{0} < x < 2x_{0} + \delta - x_{0}$ and only if $x_{0} - \delta < x < x_{0} + \delta$; hence $0 < |x| < M$ only if $|f(x+x_{0}) - A| < \varepsilon$.

Can you complete the other way around?