0
$\begingroup$

Is this correct?

If $$\lim_{x \to a} f_1(x) = L_1, \lim_{x \to a} f_2(x) = L_2, ..., \lim_{x \to a} f_n(x) = L_n$$ Then $$\lim_{x \to a} \sum_{k=1}^n f_k(x) = \sum_{k=1}^n L_k$$

Proof:

We must prove that: $$\forall \epsilon \gt 0, \exists \delta \gt 0 \; \bigl| \; 0 \lt | x - a | \lt \delta \implies |\sum_{k=1}^n f_k(x) - \sum _{k=1}^n L_k| \lt \epsilon$$

Now for any $\epsilon \gt 0$, consider $\frac \epsilon n$. Then we have: $$\delta_1 \gt 0 \; \bigl| \; 0 \lt |x - a| \lt \delta_1 \implies |f_1(x) - L_1|\lt \frac \epsilon n \\ \delta_2 \gt 0 \; \bigl| \; 0 \lt |x - a| \lt \delta_2 \implies |f_2(x) - L_2|\lt \frac \epsilon n \\ \cdot \\ \cdot \\ \cdot \\ \delta_n \gt 0 \; \bigl| \; 0 \lt |x - a| \lt \delta_n \implies |f_n(x) - L_n|\lt \frac \epsilon n$$

Considering $\delta = min\{\delta_k\}_{k = 1}^n$ we have that:

$$\delta=min\{\delta_ k\}_{k = 1}^n \; \bigl| \; 0 \lt | x - a | \lt \delta \implies \sum_{k = 1}^n|f_k(x) - L_k| \lt \sum_{k = 1}^n \frac \epsilon n = \epsilon$$

But because of the triangle inequality, it holds that: $$\sum_{k = 1}^n|f_k (x)- L_k| \ge |\sum_{k = 1}^n f_k(x) - L_k| = |\sum_{k = 1}^n f_k(x) - \sum_{k = 1}^n L_k|$$

Therefore: $$\delta=min\{\delta_ k\}_{k = 1}^n \; \bigl| \; 0 \lt | x - a | \lt \delta \implies |\sum_{k = 1}^n f_k(x) - \sum_{k = 1}^n L_k| \lt \epsilon$$

$\endgroup$
  • $\begingroup$ It's a bit superfluous to prove this for arbitrary $n$ rather than simply for $n=2$, but yes, this is the idea; it just comes down to the triangle inequality. $\endgroup$ – symplectomorphic Aug 13 '16 at 1:57
  • $\begingroup$ So is the triangle inequality in that case right? My greatest doubt was in there. A professor of mine said the symbol was supposed to be inverted, that is: $\le$ instead of $\ge$. $\endgroup$ – Carlos Afonso Aug 14 '16 at 17:11
1
$\begingroup$

Yes, this is right, though it could have been written up more carefully. For example, the notation $\delta_1\,|\,...$ doesn't mean anything. What you meant is $\color{red}{\exists}\delta_1\,|\,...$.

In the comments you've asked about your use of the triangle inequality. The key point is that

$$|f(x)+g(x)-L_1-L_2|=|f(x)-L_1+g(x)-L_2|\leq|f(x)-L_1|+|g(x)-L_2|$$

using the fact that $|a+b|\leq|a|+|b|$. The hypotheses give bounds on the righthand side, hence a bound on the lefthand side.

What you wrote is correct, but perhaps your professor read too quickly and expected you to write the inequality in the other direction. That is, you wrote (the $n$-fold generalization of)

$$|f(x)-L_1|+|g(x)-L_2|\geq|f(x)+g(x)-L_1-L_2|$$

but then you used this fact in the opposite direction. So it would have been more natural to write it the other way.

$\endgroup$

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.