Question About Limits
So one of the questions in my homework is "What is the limit as x approaches 1 of f(x)+g(x)? " So what I don't really get is that if the limit as x approaches 1 of f(x) is 3 but the limit as x approaches 1 of g(x) does not exist. Does that mean that the limit as x approaches 1 of f(x)+g(x) does not exist also?
August 16, 2012
8 Comments • Newest first
[quote=bombinator]Add the functions together first before taking the limit of it. As for your second question, I'm actually not sure, might want to ask the math tutors at the HW Help Thread
Edit: Nvm, you might want to look up your limit laws. You might want to go [url=http://dl.uncw.edu/digilib/mathematics/calculus/limits/freeze/LimitLaws.html]here[/url] and look at the "Sum Law". Basically says if both the limits of f(x) and g(x) as x approaches the same value exist, then the limit of f(x)+g(x) as x approaches that same value should also exist. So by that law, the lim of (f(x) + g(x)) as x approaches 1 should not exist.[/quote]
Thanks for all your help.
[quote=Mathematician]Okay I get it, Thanks for your help. So basically if you have to take the limits of 2 functions that are being added you add their limits to get the answer right? And a problem with no limit + 3 would be kinda non existent right?[/quote]
Add the functions together first before taking the limit of it. As for your second question, I'm actually not sure, might want to ask the math tutors at the HW Help Thread
Edit: Nvm, you might want to look up your limit laws. You might want to go [url=http://dl.uncw.edu/digilib/mathematics/calculus/limits/freeze/LimitLaws.html]here[/url] and look at the "Sum Law". Basically says if both the limits of f(x) and g(x) as x approaches the same value exist, then the limit of f(x)+g(x) as x approaches that same value should also exist. So by that law, the lim of (f(x) + g(x)) as x approaches 1 should not exist.
[quote=FallenFrost][i]The limit does not exist.[/i]
THE LIMIT DOES NOT EXIST![/quote]
Just say Limitless already.
[quote=bombinator]Yep. Try adding two functions together.
ex.
f(x) = x+1
g(x) = 1/x
If we take the lim of f(x) as x approaches infinity, it's going to be infinity.
If we take the lim of g(x) as x approaches infinity, it's going to be zero.
If we do f(x) + g(x) = (x^2+1)/x and take the limit of it as x approaches infinity, it's going to approach infinity.[/quote]
This guy knows his stuff.
But make sure to note if there are arrows, and where it's pointing.
--> and <--
[quote=BigRedMeat]>username mathematician
>can't finish his math homework
[i]ironic?[/i][/quote]
Yeah cause you know...all mathematicians sure do know every single thing about math. Also using ">" outside of that site is just plain silly.
[quote=bombinator]Yep. Try adding two functions together.
ex.
f(x) = x+1
g(x) = 1/x
If we take the lim of f(x) as x approaches infinity, it's going to be infinity.
If we take the lim of g(x) as x approaches infinity, it's going to be zero.
If we do f(x) + g(x) = (x^2+1)/x and take the limit of it as x approaches infinity, it's going to approach infinity.[/quote]
Okay I get it, Thanks for your help. So basically if you have to take the limits of 2 functions that are being added you add their limits to get the answer right? And a problem with no limit + 3 would be kinda non existent right?
I would just assume it's 3.
If the limit doesn't exist, wouldn't that just be similar to saying 3+0?
Asymptotes. That's all I remember from Pre-calculus.
Yep. Try adding two functions together.
ex.
f(x) = x+1
g(x) = 1/x
If we take the lim of f(x) as x approaches infinity, it's going to be infinity.
If we take the lim of g(x) as x approaches infinity, it's going to be zero.
If we do f(x) + g(x) = (x^2+x)/x and take the limit of it as x approaches infinity, it's going to approach infinity. I would think of it as the activity of f(x) (as x approaches infinity) "trumping" the activity of g(x).
Edit: fixed function