Determine the limit of 2^x/x^2 as x approaches infinity...

  • #1
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Homework Statement:
Determine limit of following expression as x approaches +infinity
Relevant Equations:
Limits of complex functions
How would I determine the following limit without substitution of large values of x to see what value is approached by the complex function?
## \lim_{x \rightarrow +\infty} {\dfrac {2^{x}} {x^{2} } } ## where ## x\in \mathbb{R}##
 
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  • #2
PeroK
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Use logarithms? Use L'Hopital's rule?
 
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  • #3
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Use logarithms?
I would take log of the complex expression which would still end up with infinity - infinity. Right?
 
  • #4
PeroK
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I would take log of the complex expression which would still end up with infinity - infinity. Right?
Let's see!
 
  • #5
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Let's see!
16180612061086354944016416523560.jpg
 
  • #6
PeroK
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Doesn't that help?
 
  • #9
PeroK
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Good old l'Hopital!
 
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  • #12
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Good old l'Hopital!
Can log approach be used with logic mentioned in post#11?
 
  • #13
PeroK
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So, in this case we must use L'Hopital rule.
You don't have to, but it's probably the simplest.
 
  • #14
PeroK
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We could say that x>>> ln x when x is very large, so then the last line of infinity - infinity becomes infinity, when using log approach.

Does above logic seem correct?
Yes, but you have to prove it.
 
  • #15
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Yes, but you have to prove it.
Ok, then it's not helpful. Perhaps we can use a standard formula of ## \lim_{x \rightarrow +\infty} {\dfrac {ln x} {x} } = 0 ##.

If yes then we can factor out x from second last line in my post#5 attachment, when we would get infinity ( 0- 0) which is infinity.
 
  • #16
FactChecker
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Homework Statement:: Determine limit of following expression as x approaches +infinity
Relevant Equations:: Limits of complex functions

How would I determine the following limit without substitution of large values of x to see what value is approached by the complex function?
## \lim_{x \rightarrow +\infty} {\dfrac {2^{x}} {x^{2} } }##
When you talk about a "complex function", do you mean a complicated function or do you mean a complex -valued function of a complex variable?
 
  • #17
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When you talk about a "complex function", do you mean a complicated function or do you mean a complex -valued function of a complex variable?
I meant a complicated function.
 
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  • #18
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Use logarithms? Use L'Hopital's rule?
Another possible solution could be to use Taylor series expansion of ##2^x## using the formula
as shown in screenshot below. Then it would clearly end up as (constant + infinity + infinity+...) which equals infinity.

Screenshot_20210410-200429.jpg
 
  • #19
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Homework Statement:: Determine limit of following expression as x approaches +infinity
Relevant Equations:: Limits of complex functions

How would I determine the following limit without substitution of large values of x to see what value is approached by the complex function?
## \lim_{x \rightarrow +\infty} {\dfrac {2^{x}} {x^{2} } }##
What do you mean by complex function? ##x\in \mathbb{C}## or ##x\in \mathbb{R}##? If ##x## is real you only have to figure out whether the numerator is greater than the denominator or vice versa.

But in any case: it is always a good idea to write down the definition of what must be shown.
 
  • #20
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What do you mean by complex function? ##x\in \mathbb{C}## or ##x\in \mathbb{R}##? If ##x## is real you only have to figure out whether the numerator is greater than the denominator or vice versa.

But in any case: it is always a good idea to write down the definition of what must be shown.
I didn't mean complex as in complex numbers, but an expression that is not simple. I've added that is a real number in my original question for more clarity.

Even if numerator is greater than denominator, it would be difficult to conclude the value being approached. Or I'm missing something?
 
  • #21
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I didn't mean complex as in complex numbers, but an expression that is not simple.

Even if numerator is greater than denominator, it would be difficult to conclude the value being approached. Or I'm missing something?
##2^x## grows faster than ##x^2##, which is more or less obvious. To write down the correct proof, you need to know what it means that a sequence goes to infinity. You can work with the infinity symbol, but this is not the rigorous way.
 
  • #22
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##2^x## grows faster than ##x^2##, which is more or less obvious. To write down the correct proof, you need to know what it means that a sequence goes to infinity. You can work with the infinity symbol, but this is not the rigorous way.

Would I expand ##2^x## as a Taylor series to prove the sequence going to infinity?
 
  • #23
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Would I expand ##2^x## as a Taylor series to prove the sequence going to infinity?
Begin with the definition: ##\lim_{x\to \infty } f(x)=\infty ## means, that for any ##M\in \mathbb{R}## there exists a ##C\in \mathbb{R}## such that ##f(x)>M## for all ##x > C.##

This is the formal definition. So we have to find such a ##C## which guarantees ##\dfrac{2^x}{x^2}>M## whenever ##x>C##.

We therefore transform ##\dfrac{2^x}{x^2}>M## into ##x>C## where ##C=C(M)## depends on ##M##, and write it down backwards: Given ##M\in \mathbb{R}## set ##C=\ldots## and show that ##\dfrac{2^x}{x^2}>M## for ##x> C(M).##
 
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  • #24
PeroK
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... or use L'Hopital's rule.
 
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  • #25
WWGD
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When you have a constant c with |c|>1, ##c^n## will grow much faster than ##n^c## Good for the toolbox. So limit of ratio will go to either 0 or infinity as n goes to infinity. Edit: Of course this follows and is a special case of L' Hopital, suggested by PeroK and Fresh42 in this thread.
 
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