Limit of x/(x+1) as x Goes to Infinity: Why is It 1?

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In summary, the conversation discusses the limit as x goes to infinity of x/(x+1). The participants question why the limit is 1 and not 0, and discuss using l'Hôpital's rule and factoring to compute the limit. It is explained that l'Hôpital's rule does not apply in this case and the limit is obtained by factoring and simplifying the function. The conversation ends with a discussion on the behavior of the function and its derivative at infinity.
  • #1
frasifrasi
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I am looking at the limit as x goes to infinity of x/(x+1)...

If you plug in infinity, won't it produce an indeterminate form inf/inf ? And then by taking the derivative, I get 1/(x+1)^2, which goes to 0... So can anyone explain why the limit is 1 as opposed to 0?
 
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  • #2
frasifrasi said:
And then by taking the derivative, I get 1/(x+1)^2, which goes to 0...
What does that have to do with anything?
 
  • #3
And how do you "plug in infinity" ...?
 
  • #4
it's

[tex]\frac{x}{x+1}[/tex]

not

[tex]\frac{x}{x^{-1}}[/tex]
 
  • #5
frasifrasi said:
I am looking at the limit as x goes to infinity of x/(x+1)...

If you plug in infinity, won't it produce an indeterminate form inf/inf ? And then by taking the derivative, I get 1/(x+1)^2, which goes to 0... So can anyone explain why the limit is 1 as opposed to 0?

Simply use l'Hôpital's rule to compute this limit. This indeed gives a value of 1 as x tends to infinity.
 
  • #6
since the function behaves like the y=1 line then there is nothing wrong about its derivative to be 0 at infinity.
 
  • #7
Your reference to the derivative in your first post makes me think you were trying L'Hopital's rule, but incorrectly. L'Hopital's rule says that if f(x) and g(x) both go to 0 (or to infinity)
[tex]\lim_{x\rightarrow a}= \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}[/tex]
: differentiate the numerator and denominator separately.

However, a much simpler way to handle limits at infinity, of rational functions, is to divide both numerator and denominator by the highest power of x- here just divide numerator and denominator by x. [itex]\lim_{x\rightarrow \infty} x[/itex] is difficult but [itex]\lim_{x\rightarrow \infty} 1/x[/itex] is easy!
 
  • #8
Yes, I was doing the l'hopital's rule. I understand that technique of factoring by the highest power and realized that is how the limit was obtained in the book.

But my question is, why doesn't l'hopital apply here? Doesn't the limit of the top and bottom function both go to infinity?
 
  • #9
Technique of "factoring"?
What is f and what is g, and what are their respective derivatives?
 
  • #10
Oh, i see...i was deriving the entire function rather than top and bottom...
 
  • #11
frasifrasi said:
I am looking at the limit as x goes to infinity of x/(x+1)...

If you plug in infinity, won't it produce an indeterminate form inf/inf ? And then by taking the derivative, I get 1/(x+1)^2, which goes to 0... So can anyone explain why the limit is 1 as opposed to 0?

If f(x)->a<infinity then f'(x) must go to zero. Draw a picture.

EDIT: If it is differentiable...this is not an existence inference...or maybe it is. Let me think.
 
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  • #12
[tex]\frac{x}{x+1}=1-\frac{1}{x+1}[/tex]

[tex]=1-\frac{\frac{1}{x}}{1+\frac{1}{x}}[/tex]

Now as [itex]x\rightarrow\infty[/itex],[itex]\frac{1}{x}\rightarrow0[/itex]

and therefore [tex]\frac{\frac{1}{x}}{1+\frac{1}{x}}\rightarrow0[/tex]and thus the limit is 1
 

FAQ: Limit of x/(x+1) as x Goes to Infinity: Why is It 1?

What is the limit of x/(x+1) as x goes to infinity?

The limit of x/(x+1) as x goes to infinity is 1. This means that as x gets larger and larger, the value of x/(x+1) gets closer and closer to 1.

Why does the limit of x/(x+1) as x goes to infinity approach 1?

This is because as x gets larger, the value of x+1 becomes insignificant compared to x, and therefore x/(x+1) becomes very close to 1.

Can you prove that the limit of x/(x+1) as x goes to infinity is 1?

Yes, we can use the limit definition to prove that the limit of x/(x+1) as x goes to infinity is 1. By replacing x with a very large number, we can see that the difference between x/(x+1) and 1 becomes smaller and smaller, approaching 0, which is the definition of a limit approaching 1.

Is the limit of x/(x+1) as x goes to infinity always 1?

Yes, the limit of x/(x+1) as x goes to infinity is always 1. This is a fundamental property of limits and is not affected by the specific value of x or the function x/(x+1).

Can the limit of x/(x+1) as x goes to infinity be different for different functions?

No, the limit of x/(x+1) as x goes to infinity is always 1, regardless of the function. This is because the limit is only dependent on the behavior of the function as x approaches infinity, not the specific function itself.

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