Question

prove that if the limit of f(x) as x approaches a is positive infinity and that if the limit of g(x) as x approaches a is c (where c is a real number), then the limit of [f(x)+g(x)] as x approaches a is positive infinity.

Answer 1

Well, lim x->a [f(x)+g(x)] = lim x->a f(x) + lim x->a g(x) = pos. inf + c = positive infinity. Anything added to infinity is infinity, anything subtracted from infinity is infinity. To illustrate this:

A man named Hilbert had a hotel with infinitely many rooms. Upon arriving there a guest was distressed to find that all the rooms were full. "No matter," said the manager. He asked each guest to move one room down, and there was room for the one more.

Well, the next night a carriage-load of infinitely many guests arrived. But again the hotel was full! But no worries. The manager, rubbing his hands in anticipation of infinitely many hotel bills, asked each guest to move down 2x rooms, where x = the room number they were in. In this way, every guest was accommodated. This is a basic illustration of infinity, colloquially known as "Hilbert's Hotel."

Answer 2

Thank you Tangent. However, this explanation is solely based on intuition. What I am after is the proof of the problem using the proper definition of the limits, infinite limits etc. I get your point. Thanks anyway.

Answer 3

what definition are you using for the limit?

Answer 4

we are using the epsilon-delta definition.

Answer 5

usually to say that
\[\lim_{x->\infty}f(x)=\infty\] means given any N you can find an M such that if
\[x>M\] then
\[f(x)>N\]

Answer 6

Yes, that is where I am stuck right now. I don't know how to incorporate that into the problem. :c

Answer 7

oh wait i am sorry. you have
\[\lim_{x->a}f(x)=\infty\]

Answer 8

means given any N you can find
\[\delta=\delta(N)\] such that if
\[|x-a|<\delta\] then
\[f(x)>N\]

Answer 9

first of all it is entirely clear that
\[\lim_{x->a}(f(x)+g(x)=\lim_{x->a}f(x)+\lim_{x->a}g(x)
\] yes?

Answer 10

and you want to say that this is infinite. this is the same as saying
\[\lim_{x->a}f(x)+c\] is infinite, since the second limit is c

Answer 11

so you need to say that given any N > 0 there is a
\[\delta\] such that if
\[|x-a|<\delta\] then
\[f(x)+c > N\] meaning of course that
\[f(x)> N-c\]

Answer 12

but you have that the limit as x goes to a of f is infinite, so you can pick the delta that works for N - c and you are done

Answer 13

that is, say that since
\[\lim_{x->a}f(x)=\infty\] there is a
\[\delta\] such that if
\[|x-a|<\delta\] then
\[f(x)> N-c\] and you are done

Answer 14

btw it is very hard to get a drink at the hilbert hotel.

Answer 15

thank you so much. it is really a hell of a time in that hotel, with all those shizz of people. :)