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I was wondering why the statement in the title is true only if the functions we are dealing with are continuous.

Here's the context (perhaps not required):

enter image description here

(The upper equation there is just a limit of two sums, and the lower expression is two limits of those two sums.), and if anyone wonders, that's the original source (a pdf explaining the proof of the product rule).

P.S. In the context it's given that $g$ and $f$ are differentiable, anyway I only provided it to illustrate the question; my actual question is simply general.

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It's not true that the limit of a sum is the sum of the limits only if the functions are continuous. It is true if the latter limits exist. Such is the case if you're dealing with continuous functions. –  David Mitra Sep 12 '12 at 16:14

3 Answers 3

up vote 1 down vote accepted

Take $$f(x)=\cases{0 & for $x<0$\\ 1 & for $x \ge 0$}$$ and $g(x)=1-f(x)$. Note that both $f$ and $g$ are discontinuous at $x=0$. Then $\lim_{x\to 0}(f(x)+g(x))=\lim_{x\to 0}1 = 1$, but neither $\lim_{x\to 0}f(x)$ nor $\lim_{x\to 0}g(x)$ exists, and thus also not their sum.

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@DavidMitra: Yes. If the limits don't exist, the sum of the limits doesn't exist either. –  celtschk Sep 12 '12 at 16:23

The algebraic limit theorem(s) tell you that if you have a limit of a finite sum of functions, and the limit of each function exists then the limit of the sum is the sum of the limits. This goes for products and quotients as well (assuming in the latter case that the denominator does not converge to 0).

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The limit of the sum is not always equal to the sum of the limits, even when the individual limits exist.

For example:

Define $h(i)=1\sqrt{(n^2)+i}$. For each $i=1,\cdots,n$, the limit of $h(i)$ is zero as n goes to infinity.

But the limit of the sum $[h(1)+h(2)+\cdots+h(n)]$ as n goes infinity is not zero.

The limit of this sum is actually $1$, using the sandwich theorem.

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