- #1
Tsunoyukami
- 215
- 11
Prove the Sum Rule for Limits
$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$
Proof
Assume the following:
$$\lim_{x \to a} f(x) = L, \space\lim_{x \to a} g(x) = M$$
Then, by definition
##\forall \epsilon_1 > 0, \exists \delta_1 > 0## such that ##0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon_1##
and
##\forall \epsilon_2 >0, \exists \delta_2 > 0## such that ##0<|x-a|<\delta_2 \implies |g(x)-M|<\epsilon_2##
Choose ##\delta = \min(\delta_1, \delta_2)##.
Then ##0<|x-a|<\delta \implies |f(x)-L|<\epsilon_1## and ##|g(x)-M|<\epsilon_2##.
Notice ##|[f(x) + g(x)] - [L+M]| = |[f(x)-L] + [g(x)-M]| \leq |f(x)-L| + |g(x)-M|## by the triangle inequality. Then ##|[f(x) + g(x)] - [L+M]| \leq |f(x)-L| + |g(x)-M| < \epsilon_1 + \epsilon_2 = \epsilon##.
Then ##|[f(x) + g(x)] - [L+M]| < \epsilon##. Therefore we conclude
$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$
I was comparing my proof (the one above) to the proof offered on page 93 of James Stewart's Calculus (6e) and noticed that instead of ##\epsilon_1## and ##\epsilon_2## he uses ##\frac{\epsilon}{2}## for each of these so that their sum is ##\epsilon##.
Is there anything incorrect with my method? I figure that my approach with two different possible values of ##\epsilon_i## for each of the limits is just a more generalized way of saying the same thing - but the other proofs I've looked at online all use ##\frac{\epsilon}{2}## - is there any reason for this?
Thanks in advance!
$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$
Proof
Assume the following:
$$\lim_{x \to a} f(x) = L, \space\lim_{x \to a} g(x) = M$$
Then, by definition
##\forall \epsilon_1 > 0, \exists \delta_1 > 0## such that ##0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon_1##
and
##\forall \epsilon_2 >0, \exists \delta_2 > 0## such that ##0<|x-a|<\delta_2 \implies |g(x)-M|<\epsilon_2##
Choose ##\delta = \min(\delta_1, \delta_2)##.
Then ##0<|x-a|<\delta \implies |f(x)-L|<\epsilon_1## and ##|g(x)-M|<\epsilon_2##.
Notice ##|[f(x) + g(x)] - [L+M]| = |[f(x)-L] + [g(x)-M]| \leq |f(x)-L| + |g(x)-M|## by the triangle inequality. Then ##|[f(x) + g(x)] - [L+M]| \leq |f(x)-L| + |g(x)-M| < \epsilon_1 + \epsilon_2 = \epsilon##.
Then ##|[f(x) + g(x)] - [L+M]| < \epsilon##. Therefore we conclude
$$\lim_{x\to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$$
I was comparing my proof (the one above) to the proof offered on page 93 of James Stewart's Calculus (6e) and noticed that instead of ##\epsilon_1## and ##\epsilon_2## he uses ##\frac{\epsilon}{2}## for each of these so that their sum is ##\epsilon##.
Is there anything incorrect with my method? I figure that my approach with two different possible values of ##\epsilon_i## for each of the limits is just a more generalized way of saying the same thing - but the other proofs I've looked at online all use ##\frac{\epsilon}{2}## - is there any reason for this?
Thanks in advance!