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Please get me started on showing that the following limit exists

by kingstrick
Tags: exists, limit, showing, started
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kingstrick
#1
Mar14-12, 05:41 PM
P: 108
1. The problem statement, all variables and given/known data

Evaluate the limit or show it doesn't exist. (x→∞) lim ((x+2)/√x) where (x > 0)

2. Relevant equations



3. The attempt at a solution

I know how to solve it if x → c but i don't know how to start it when it goes to infinity. I just need a hint as to how to start the problem.
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Mark44
#2
Mar14-12, 06:02 PM
Mentor
P: 21,397
Quote Quote by kingstrick View Post
1. The problem statement, all variables and given/known data

Evaluate the limit or show it doesn't exist. (x→∞) lim ((x+2)/√x) where (x > 0)

2. Relevant equations



3. The attempt at a solution

I know how to solve it if x → c but i don't know how to start it when it goes to infinity. I just need a hint as to how to start the problem.
Divide each term in the numerator by √x, and then take the limit.
kingstrick
#3
Mar14-12, 09:34 PM
P: 108
Quote Quote by Mark44 View Post
Divide each term in the numerator by √x, and then take the limit.
Mark, thanks for responding,

so after evaluating, i found that the limit does not exists... am i correct?
work:
((x+2)/√x)/1 --- Does not exist since the numerator is always bigger than the denominator
x→∞, x > o

scurty
#4
Mar14-12, 10:00 PM
P: 392
Please get me started on showing that the following limit exists

Quote Quote by kingstrick View Post
Mark, thanks for responding,

so after evaluating, i found that the limit does not exists... am i correct?
work:
((x+2)/√x)/1 --- Does not exist since the numerator is always bigger than the denominator
x→∞, x > o
That's not what Mark meant. You essentially have the exact same equation you started with. Remember that

[itex]\displaystyle\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}[/itex].
Mark44
#5
Mar14-12, 11:54 PM
Mentor
P: 21,397
Quote Quote by kingstrick View Post
so after evaluating, i found that the limit does not exists... am i correct?
work:
((x+2)/√x)/1 --- Does not exist since the numerator is always bigger than the denominator
x→∞, x > o
In one sense, the limit doesn't exist, because (x + 2)/√x gets large without bound as x gets large. In that sense, the limit doesn't exist because it is not a finite number. For limits like this, though, we usually say that the limit is ∞.

Also, because x is approaching infinity, you don't need to say that x > 0.
kingstrick
#6
Mar15-12, 07:13 AM
P: 108
Quote Quote by Mark44 View Post
In one sense, the limit doesn't exist, because (x + 2)/√x gets large without bound as x gets large. In that sense, the limit doesn't exist because it is not a finite number. For limits like this, though, we usually say that the limit is ∞.

Also, because x is approaching infinity, you don't need to say that x > 0.
So when x → ∞, a function will always either go to a finite number, ∞, or -∞...so will a function then always have a limit in this sense? Oh, nevermind, some functions can diverge, like f(x) = -1^x.
Mark44
#7
Mar15-12, 08:33 AM
Mentor
P: 21,397
Quote Quote by kingstrick View Post
So when x → ∞, a function will always either go to a finite number, ∞, or -∞...so will a function then always have a limit in this sense? Oh, nevermind, some functions can diverge, like f(x) = -1^x.
Right, except that ##\lim_{x \to \infty}-1^x = -1##

The one you're thinking of is f(x) = (-1)x. Without parentheses, what you wrote is the same as -(1x).
kingstrick
#8
Mar15-12, 09:05 AM
P: 108
Thank you. I think i understand now.


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