# How to prove $\lim_{x\to a}f(x) = \lim_{x\to b}g(x)$

For example, I have the following question in a book:

Prove that $$\lim_{x\to 0}f(x) = \lim_{x\to a}f(x-a).$$

To be clear, I am just finishing a chapter about limits and do not know anything about topics beyond that in calculus, such as derivatives, integrals, sequences etc. I know about the definition of a limit of a function, the additive, multiplicative and divisive theorems about limits and the theorem on uniqueness of limits.

Since I have never been shown an equality between two limits in this way before, I started contemplating how I should begin. My initial method was to assume that $\lim_{x\to 0}f(x) = L_1$ and that $\lim_{x\to a}f(x-a) = L_2$ and then prove that $L_1 = L_2$. But then it struck me that this must be an inconclusive method because any one of the two limits might not exist, so I need to prove that $L_1$ exists iff $L_2$ exists, apart from $L_1 = L_2$. Is this correct?

Then, I tried to work with the definition of a limit. I set $f(x-a) = g(x)$ and tried to prove $$\forall \epsilon > 0 \quad \exists \delta > 0 \quad ( \forall x\in D_f \quad(|x|<\delta \implies |f(x) - L|<\epsilon)) \iff \\ \forall \epsilon_2 > 0 \quad \exists \delta_2 > 0 \quad( \forall x\in D_g \quad(|x-a|<\delta_2 \implies |g(x) - L|<\epsilon_2))$$

Now it all seems to be pretty straightforward because $D_f$ and $D_g$ has the same amount of elements and for every $x$ in $D_f$, $(x-a)\in D_g$ and for every $x$ in $D_g$, $(x+a)\in D_f$. Therefore it is easy to plug in $(x-a)$, $D_g$ in the left hand side and $(x+a)$, $D_f$ in the right hand side which shows the equivalence. I feel a bit insecure though, is this a valid proof strategy?

As a side note, the answer just made me even more confused:

Suppose $\lim_{x\to a} f(x) = L$, and let $g(x) = f(x-a)$. Then for all $\epsilon > 0$ there is a $\delta > 0$ such that, for all $x$, if $0 < |x-a| < \delta$, then $|f(x) - L| < \epsilon$. Now, if $0 < |y| < \delta$, then $0 < |(y + a)-a| < \delta$, so $|f(y+a) - L| < \epsilon$. But this last inequality can be written $|g(y) - L| < \epsilon$. So $\lim_{y\to 0}g(y) = L$. The argument in the reverse direction is similar.

I feel like this proof makes no sense with regard to the question, because I feel like he has proven that if $\lim_{x\to a}f(x) = L$ then $\lim_{y \to 0}g(y) = L$, while the questions asks for equality between $\lim_{x\to 0}f(x)$ and $\lim_{y\to a}f(y-a) = \lim_{y\to a}g(y)$, right? So he should start by assuming $\lim_{x\to 0}f(x) = L$, am I right?

• The answer/proof of your book is perfectly correct and works by the definition. What part is bothering you, I could give an explanation. :) May 4 '16 at 11:24
• @CharalamposFilippatos See my edit, ask if you need further clarification :) May 4 '16 at 11:34
• Sorry for answering some minutes later. I gave you an analytic answer down below (edited it some times). I hope you will understand it, give it a look ! May 4 '16 at 11:52

Remember that your $y$ there is : $y = x-a$, so for $x\rightarrow 0 \Leftrightarrow y \rightarrow -a$ . He shows, by using the definition of the limit, that if $\lim_{x \to 0} f(x) = L$ then $\lim_{y \to -a} g(y) = L$. This is exactly shown the way he proceeds, by starting from what you have and building the new expression by definition.
Suppose that $\lim_{x \to 0} f(x) = L$ and that $g(x) = f(x-a)$. By the limit definition then, $\forall ε>0$ $\exists δ>0 : \forall x$ if $0<|x-0|<δ$ then $|f(x)-L|<ε$. This is the definition of : $\lim_{x \to 0} f(x) = L$. But if the $y$ that you change is : $0<|y|<δ$ (note that $y=x-a$ , then you will have that : $0<|(y-a)+a|<δ$ to keep y intact. By that, you use the definition, and it goes $|f(y-a) - L|< ε$. But since you proved by definition that $\lim_{x \to a} f(x) = L$, this means that : $|g(y) - L|<ε$ which means that $\lim_{y \to o} g(y) = L$. And this gives you : $\lim_{x \to a} g(x) = L$ (because when $y->0$ you have that : $x-> a$
• Thank you for your answer. I understand it better now but can we really assume that $x \to a \iff y \to 0$? I thought this would follow If we could prove the original statement :P Anyways, I will accept this as an answer and think about this a little bit more and then hopefully I will understand fully! May 4 '16 at 12:06
• Yes, you can assume that. Let $h(x) = x$ and $t(x) = h(x) - a$. Then, $\lim_{x \to a} t(x) = 0$ May 4 '16 at 12:21
• I just have to ask again because I can't seem to figure this out: You start with the assumption that $\lim_{x\to a}f(x) = L$ and then prove $\lim_{y\to 0}g(y) = L$. But shouldn't the proof be that $\lim_{x\to 0}f(x) = L \iff \lim_{y\to 0}g(y) = L$? Why do you use $\lim_{x\to a}f(x) = L$ in the beginning of your proof instead of $\lim_{x\to 0}f(x) = L$? After all, the equality in the question does not even involve $\lim_{x\to a}f(x)$, does it? May 4 '16 at 14:07