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How would I go about proving that two limits are equal to each other using the Epsilon-Delta definition?

Moreover how can I prove that: $$\lim_{x\to0}f(x) = \lim_{x\to a}f(x-a)$$ using the Epsilon-Delta definition? The intuition for this seem clear. However, I have do not know how a formal proof can be developed.

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  • $\begingroup$ If the two limits are unrelated, you have no other option than computing them (by the $\epsilon/\delta$ definition) and comparing the values. $\endgroup$ – Yves Daoust Nov 23 '18 at 17:31
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To show at least the second problem, assume that $\lim_{x\to 0} f(x) = L$. We will prove that $\lim_{x \to a}f(x-a) = L$.

By definition, we are to show : for all $\epsilon > 0$, there exists $\delta > 0$ such that \begin{equation} |y - a |< \delta \implies |f(y-a) - L| < \epsilon \tag{1} \end{equation}

Fix one such $\epsilon$, say $\epsilon_0$.

We know that $\lim_{x \to 0} f(x) = L$. This means, that for the above $\epsilon_0$, there exists some $\delta_0 > 0$ such that \begin{equation}|x - 0|( = |x|) < \delta_0 \implies |f(x) - L| < \epsilon_0 \tag{2}\end{equation}

By taking $\delta = \delta_0$, we see that substituting $y-a = x$ in $(2)$ gives: \begin{equation} |y-a| < \delta_0 \implies |f(y-a) - L |< \epsilon_0 \end{equation}

which is seen to be equivalent to statement $(1)$. Hence, $\lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)$ is true.

I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $\lim_{x \to a} f(x) \neq L$ for some constant $L$.

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  1. express the first limit by the $\epsilon/\delta$ definition.

  2. formally replace every occurrence of $x$ by $x-a$.

  3. now you have the proof of the second limit.


I.e. rewrite

$$\lim_{x\to 0}f(x)=L \\\iff \\\forall\epsilon>0:\exists\delta>0:|x|<\delta\implies|f(x)-L|<\epsilon \\\iff \\\forall\epsilon>0:\exists\delta>0:|x-a|<\delta\implies|f(x-a)-L|<\epsilon \\\iff\\\lim_{x-a\to 0}f(x-a)=L. $$

The argument holds because $x$ and $x-a$ cover the same set of values, and $\lim_{x-a\to0}$ is also $\lim_{x\to a}$.

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