If $\lim_{x \to c}f(x) = L$ show that $\lim_{x \to 0}f(x+c) = L$

Question

I've been at this question for some time now. I tried showing $$\lim_{x \to 0}f(x+c)$$ can be rewritten as $$\lim_{x \to c}f(x),$$ but I couldn't show that either.

Anybody know what I can do?

Answer 1

If $$\lim _{ x\rightarrow c }{ f\left( x \right) =L } \Rightarrow \forall \epsilon >0,\exists \delta \left( \epsilon \right) >0\quad \left| x-c \right| <\delta \left( \epsilon \right) ,\left| f\left( x \right) -L \right| <\epsilon $$ Now let $x-c=t$ so that $$x\rightarrow c\quad \Rightarrow t\rightarrow 0$$ by substitution we get $x=t+c$

$$\forall \epsilon >0,\exists \delta \left( \epsilon \right) >0\quad \left| t \right| <\delta \left( \epsilon \right) ,\left| f\left( t+c \right) -L \right| <\epsilon $$ which means $$\lim _{ t\rightarrow 0 }{ f\left( t+c \right) =L } $$

Answer 2

You could see at the limit of the composite function to solve the question or get an hint

Answer 3

Indeed this is a particular case of the change of variable rule: suppose that $f(x)\rightarrow A$ when $x\rightarrow x_0$ and $g(y)\rightarrow x_0$ when $y\rightarrow y_0$ then

$$\lim_{y\rightarrow y_0} f(g(y))=A $$

and a proof can goes as follow:

For each $\epsilon>0$ you can find $\delta'>0$ such that if $|x-x_0|<\delta'$ then $|f(x)-A|<\epsilon$.

And for this $\delta'$ you can find a $\delta>0$ such that if $|y-y_0|<\delta$ then $|g(y)-x_0|<\delta'$ but then $|f(g(y))-A|<\epsilon$.

Then

$$\lim_{y\rightarrow y_0} f(g(y))=A $$

If you only want that for the linera case you can use exactly the same argument. But in that case you have an advantage since in the proof you can use exactly the same $\delta$.

Answer 4

I'll try to rephrase haqnatural's answer a bit:

Given:. $\lim_ {y \rightarrow c} f(y) = L.$

For $\epsilon \gt 0,$ exists a $\delta \gt 0$ such that

$|y-c| \lt \delta$ implies $|f(y) -L| \lt \epsilon.$

Set $x:= y-c,$ then $ |x| = |y-c| \lt \delta$, which implies

$|f(x+c) - L| = |f(y) -L| \lt \epsilon$.

In terms of $x$ only:

$|x| \lt \delta$ implies $ |f(x+c) -L| \lt \epsilon$, I.e.

$\lim_{x \rightarrow 0} f(x+c)=L.$