A small doubt in a limit problem

[basheer uddin]

Homework Statement

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

Homework Equations

l'hospital's rule

The Attempt at a Solution

on substitution we get
$\frac { \infty }{ \infty }$
on using l'hospital's rule,what is the result of differentiation of x!?
or should we even use l'hopital's rule?

 

[ehild]

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

ehild

 

[HallsofIvy]

You can't use L'Hopital's rule because n! is not a differentiable function.

 

[basheer uddin]

if we can't use l'hopital's rule then which method should we use?

 

[basheer uddin]

on plotting i got that it initially approaches ∞ then it goes down to zero.but how do we solve it mathematically?

 

[Ray Vickson]

if we can't use l'hopital's rule then which method should we use?

Look up "Stirling's Formula".

 

[Curious3141]

Homework Statement

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

Homework Equations

l'hospital's rule

The Attempt at a Solution

on substitution we get
$\frac { \infty }{ \infty }$
on using l'hospital's rule,what is the result of differentiation of x!?
or should we even use l'hopital's rule?

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 $\lim_{n\to \infty} a_n$?

 

[pbuk]

What is the ratio of successive terms?

 

[HallsofIvy]

What is the ratio of successive terms?

basheer udden, this is an excellent suggestion! Two consecutive terms are \frac{10^x}{x!} and \frac{10^{x+ 1}}{(x+1)!}. What is the first divided by the second? What does that tell you?

 

[basheer uddin]

basheer udden, this is an excellent suggestion! Two consecutive terms are \frac{10^x}{x!} and \frac{10^{x+ 1}}{(x+1)!}. What is the first divided by the second? What does that tell you?

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?

 

[ehild]

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

 

[pbuk]

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.