0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ The answer/proof of your book is perfectly correct and works by the definition. What part is bothering you, I could give an explanation. :) $\endgroup$
    – Rebellos
    May 4 '16 at 11:24
  • $\begingroup$ @CharalamposFilippatos See my edit, ask if you need further clarification :) $\endgroup$ May 4 '16 at 11:34
  • $\begingroup$ 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 ! $\endgroup$
    – Rebellos
    May 4 '16 at 11:52
2
$\begingroup$

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 : $x-> a$

$\endgroup$
6
  • $\begingroup$ 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! $\endgroup$ May 4 '16 at 12:06
  • $\begingroup$ Yes, you can assume that. Let $ h(x) = x $ and $t(x) = h(x) - a$. Then, $ \lim_{x \to a} t(x) = 0 $ $\endgroup$
    – Rebellos
    May 4 '16 at 12:21
  • $\begingroup$ 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? $\endgroup$ May 4 '16 at 14:07
  • $\begingroup$ Check my new answer. It indeed seemed weird in the second look, like being another exercise, I didn't pay attention to the starting phrase but only at the proof. Anyway, check my answer now. $\endgroup$
    – Rebellos
    May 4 '16 at 14:42
  • $\begingroup$ Check that answer, it's the correct one. $\endgroup$
    – Rebellos
    May 4 '16 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.