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A small doubt in a limit problem

  1. Aug 23, 2014 #1
    1. The problem statement, all variables and given/known data

    find limit
    ##\lim _{ x\rightarrow \infty }{ \frac { { 10 }^{ x } }{ x! } } ##

    2. Relevant equations

    l'hopitals rule

    3. The attempt at a solution
    on substitution we get
    ##\frac { \infty }{ \infty } ##
    on using l'hopitals rule,what is the result of differentiation of x!?
    or should we even use l'hopital's rule?
     
  2. jcsd
  3. Aug 23, 2014 #2

    ehild

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    Can you find the limit of the sequence ##a_n=\frac{10^n}{n!}?##

    ehild
     
  4. Aug 23, 2014 #3

    HallsofIvy

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    You can't use L'Hopital's rule because n! is not a differentiable function.
     
  5. Aug 23, 2014 #4
    if we can't use l'hopital's rule then which method should we use?
     
  6. Aug 23, 2014 #5
    on plotting i got that it initially approaches ∞ then it goes down to zero.but how do we solve it mathematically?
     

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  7. Aug 23, 2014 #6

    Ray Vickson

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    Look up "Stirling's Formula".
     
  8. Aug 23, 2014 #7

    Curious3141

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    One way is to use Stirling's approximation, as Ray Vickson has advised. This will allow you to approximate the denominator by an analytic function and you can then use L' Hopital's rule.

    Another way is to simply quote a well known Maclaurin series that, for a particular argument, will have exactly ##a_n=\frac{10^n}{n!}## as its general term. Since you know that series to be convergent, what can you say about ##\lim_{n\to \infty} a_n##?
     
  9. Aug 23, 2014 #8
    What is the ratio of successive terms?
     
  10. Aug 23, 2014 #9

    HallsofIvy

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    basheer udden, this is an excellent suggestion! Two consecutive terms are [itex]\frac{10^x}{x!}[/itex] and [itex]\frac{10^{x+ 1}}{(x+1)!}[/itex]. What is the first divided by the second? What does that tell you?
     
  11. Aug 24, 2014 #10
    well the ratio of successive terms is
    ##\frac { 10 }{ x+1 } ##
    and for x>9 the ratio is <1.so that means, as x→∞ the ratio tends to 0.so if we continue multiplying the equation with ##\frac { 10 }{ x+1 } ## ultimately it tends to zero so the limit is zero!
    is my logic correct?
     
  12. Aug 24, 2014 #11

    ehild

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    Does x mean n, 1, 2, 3, ... ? So is f(x) a sequence? If it is a function, you can not speak about consecutive terms.

    If it is a sequence, you are right. It would tend to zero even in the case the ratio was less than 1 in absolute value.


    ehild
     
  13. Aug 24, 2014 #12
    Well if it is a function its domain is the non-negative integers (unless the question is using a non-standard definition of !) so the range of the function is identical to the set of terms of the sequence which you yourself mentioned ehild. The question was "find the limit" not "prove that the limit is..." so I don't think the answer needs to make a big deal about using this equivalence.

    Basheer uddin your logic is correct.
     
    Last edited: Aug 24, 2014
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