Why does Lim as x approaches infinity of x/(x-9) = 1?

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In summary, the limit as x approaches infinity of x/x-9 can be evaluated by dividing both the numerator and denominator by x, simplifying, and then applying L'Hopital's rule to get a final result of 1/1 or simply 1. This is due to the fact that as x gets larger and larger, the discrepancy between x and x-9 diminishes, resulting in an approaching value of 1.
  • #1
LOLItsAJ
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Homework Statement



Limit as x approaches infinity of [itex]x/x-9[/itex]

Homework Equations



None

The Attempt at a Solution



I know the indeterminate form infinity/infinity happens. I don't know how to fix it, but I'm assuming it's quite simple...
 
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  • #2
When x is very very large, say 6 billion. The numbers 6 billion and 6 billion minus 9 are essentially the same. As x gets larger and larger, this discrepancy diminishes. The value the ratio approaches is 1.

If you wanted to be a little more rigorous, you can divide the numerator and denominator by 1/x then take the limit and see what happens. It should pop right out.
 
  • #3
A nice trick for these types of problems would be to divide everything by x;

[tex] \lim_{x\rightarrow\infty} \frac{x/x}{(x/x)-(9/x)} [/tex]

Then simplify and you should be able to figure it out from there
 
  • #4
Using L'Hopitals rule for evaluating indeterminate forms, just differentiate both the numerator and the denominator separately to produce 1/1. Xyius explains well intuitively why the result is what it is.
 

Related to Why does Lim as x approaches infinity of x/(x-9) = 1?

What is the limit as x approaches infinity of x/(x-9)?

The limit as x approaches infinity of x/(x-9) is equal to 1.

Why does the limit as x approaches infinity of x/(x-9) equal 1?

This can be explained by the fact that as x approaches infinity, the value of x/(x-9) also approaches 1. This is because as x gets larger and larger, the denominator (x-9) becomes insignificant in comparison to the numerator x.

How can I prove that the limit as x approaches infinity of x/(x-9) is equal to 1?

One way to prove this is by using the limit definition and showing that for any small number ε, there exists a large enough value of x such that the difference between x/(x-9) and 1 is smaller than ε.

What happens if I approach infinity from the other direction, i.e. x approaches negative infinity?

The result would still be the same, as both x and (x-9) would approach negative infinity. Therefore, the limit as x approaches negative infinity of x/(x-9) would also be equal to 1.

Are there any other values that the limit as x approaches infinity of x/(x-9) could be?

No, the limit as x approaches infinity of x/(x-9) is only equal to 1 as x approaches positive or negative infinity. This is because the function x/(x-9) has a horizontal asymptote at y=1 as x approaches infinity.

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