how do I prove that:
lim (f(x)+g(x)) = lim f(x) + lim g(x)
x->a x->a x->a
?
#Limit formula proof
trim moon
bright orchidBOT
how do I prove that:
lim (f(x)+g(x)) = lim f(x) + lim g(x)
x->a ...
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trim moon
can u latex that
idk how
wide ridge
can u latex that
Show that $$\lim\limits_{x\to a}\big(f(x)+g(x)\big) = \lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x)$$
prime steppeBOT
toe
prime steppeBOT
wolf
wide ridge
Recall the definition of the derivative by first principles
Maybe that might work?
idk actually, usually I see this as just an axiom of limits
trim moon
Recall the definition of the derivative by first principles
derivative formulas are derived from limits
wide ridge
yes
trim moon
so I want to prove it without derivatives
wide ridge
tbh I don't know whether the statement is true
tacit sinew
the statement is perfectly true
wide ridge
Let $a = 0$, $f(x) = \frac x{|x|}$ and $g(x) = -\frac x{|x|}$
tacit sinew
hmm
trim moon
Let $a = 0$, $f(x) = \frac x{|x|}$ and $g(x) = -\frac x{|x|}$
so it's only true for when both f(x) and g(x) exists at x=a? but still how do you prove it?
wide ridge
This might be the statement, but idk
Perhaps we can use the epsilon-delta definition of limits? I'm not actually sure how to prove the statement
trim moon
Perhaps we can use the epsilon-delta definition of limits? I'm not actually sure...
I ever heard of it but idk what it is
gaunt nexus
Do it graphically
trim moon
Do it graphically
I ever tried to imagine it graphically but not sure if it'll work
trim moon
Hoe about we use the definitions ?
how?
wind bough
Well the definition of lim as x -> a of f(x) = F is just a shorthand notation for the proposition:
For any epsilon > 0, there exists a delta > 0 such that for any x that |x-a| < delta then it's true that |f(x) - F| < epsilon
The same goes for *lim as x -> a of g(x) = G
For any epsilon > 0, there exists a delta > 0 such that for any x that |x-a| < delta then it's true that |g(x) - G| < epsilon
Now just focus on the last part of both of these 2 propositions
|f(x) - F| < epsilon and |g(x) - G| < epsilon
Think you can combine these 2 things in such a way that will give us something about f(x) + g(x) - (F + G) ?
PS: We want to do this thing because we want to prove in the end that |f(x) + g(x) - (F + G)| < epsilon i.e. the fact that lim x->a of f(x) + g(x) = F + G , which is our objective
Hint: ||Since we are working with inequliaties and absolute values, wouldn't it be useful to use the triangle inequality ( |a| + |b| >= |a+b| ) somehow?||
Finalising the proof:
|| Since |f(x) - F| < epsilon and |g(x) - G| < epsilon then their sum |f(x) - F| + |g(x) - G| > |f(x) - F + g(x) - G| = |f(x) + g(x) - (F+G)|
Since these 2 things are smaller than their epsilon then |f(x) + g(x) - (F+G)| < epsilon1 + epsilon2, where we will call epsilon1 + epsilon2 = epsilon3 from now on (note that epsilon >0 since epsilon1 and epsilon2 are > 0, so there aren't any problems here)
So now we need to find a delta such that any x that |x-a| < delta will give us |f(x) + g(x) - (F+G)| < epsilon3
It may take some thinking but a good choice for this delta is min{delta1, delta2}, where delta1 and delta2 are the deltas associated for epsilon1 and epsilon2 from our premises.
Let's check if it's a good choice:
We pick an epsilon3, then get the delta = min{delta1, delta2}
This branches into 2 symmetric parts:
If delta1 < delta2: delta = delta1 then it means that |x-a| < delta1 < delta 2 which means it's obvious that |x-a| < delta1 is true and also |x-a| < delta2 is also true so it means that |f(x) - F| < epsilon3 and |g(x) - g| < epsilon3 is also true and so |f(x) + g(x) - (F + G)| < epsilon3 is obvious as well.
The other branch where delta2 < delta1 goes the same way so there aren't any problems.
So in the end we proved that lim x -> a of f(x) + g(x) = F+G = lim x->a of f(x) + lim x->a of g(x)
||