# Limit of x^α.sin²(x!)/(x+1) as x approaches infinity is?

## Homework Statement

:
Find [/B]
limx->∞(xα(sin2x!)/(x+1)
α∈(0,1)
Options are:
a)0
b)1
c)inifinity
d)does not exist

: -[/B]

## The Attempt at a Solution

:
limx->∞(xαsin2x!)/(x+1)[/B]
Dividing the numerator and denominator by xα,
we have:limx->∞sin2x!)/(x1-α+x)
clearly x−αis tending to zero as x tends to infinity
and thus we have 1/x1-α tending to zero
[as x tends to infimity, we can say the denomimator tends to x1-α ]
thus we have (a number tending to zero)*(a sinusoidal function which is largely changing values as x tends to infinity)
so i feel limit does not exist because of that oscillating part.
My friends feel answer is zero as the sin part has a finite value, so we have
(tending to zero)*(finite number) which gives tending to zero.

And THE ANSWER GIVEN IN TEXT IS INFINITY.
Help plz.

Last edited:

Mark44
Mentor

## Homework Statement

:
Find [/B]
limx->0(xαsin2x!)/(x+1)
α∈(0,1)
Are you sure this is the exact problem statement? I'm bothered by the expression ##\sin^2(x!)##. The factorial is typically applied to nonnegative integer values, which is not what you have in this limit problem. There is something called the Gamma function (##\Gamma## function) that is related to factorials, without the restrict that the arguments have to be integers >= 0.

If this is the exact statement, I'm wondering if the above is a typo.
BTW, is the limit as ##x \to \infty## or as ##x \to 0##?
randomgamernerd said:
Options are:
a)0
b)1
c)inifinity
d)does not exist

: -[/B]

## The Attempt at a Solution

:
limx->0(xαsin2x!)/(x+1)[/B]
Dividing the numerator and denominator by xα,
we have:limx->0sin2x!)/(x1-α+x)
clearly x−αis tending to zero as x tends to infinity
and thus we have 1/x1-α tending to zero
[as x tends to infimity, we can say the denomimator tends to x1-α ]
thus we have (a number tending to zero)*(a sinusoidal function which is largely changing values as x tends to infinity)
so i feel limit does not exist because of that oscillating part.
My friends feel answer is zero as the sin part has a finite value, so we have
(tending to zero)*(finite number) which gives tending to zero.

And THE ANSWER GIVEN IN TEXT IS INFINITY.
Help plz.

i
Are you sure this is the exact problem statement? I'm bothered by the expression ##\sin^2(x!)##. The factorial is typically applied to nonnegative integer values, which is not what you have in this limit problem. There is something called the Gamma function (##\Gamma## function) that is related to factorials, without the restrict that the arguments have to be integers >= 0.

If this is the exact statement, I'm wondering if the above is a typo.
BTW, is the limit as ##x \to \infty## or as ##x \to 0##?
I'm so sorry, I've made a typo..its limit x tends to infinity, not zero

Mark44
Mentor
I agree with your friends who are saying the limit is zero. If you split it into ##\frac{x^\alpha}{x + 1} \sin^2(x!)##, the first factor is going to zero, and the second factor is bounded between 0 and 1.

I agree with your friends who are saying the limit is zero. If you split it into ##\frac{x^\alpha}{x + 1} \sin^2(x!)##, the first factor is going to zero, and the second factor is bounded between 0 and 1.
so, the answer is no way infinity, right?
and can you please upload a graph of the function? i am unable to construct it though i used online graphing calculators

epenguin
Homework Helper
Gold Member
I agree with your friends who are saying the limit is zero. If you split it into ##\frac{x^\alpha}{x + 1} \sin^2(x!)##, the first factor is going to zero, and the second factor is bounded between 0 and 1.
The number of ( in a problem should equal the number of ) which don't think they do at the moment.#1 Well not in the first line, I see it is OK further down.

Also, Mark44, doesn't that depend whether |α| ≥ 1 or not?

I think you can talk about limits of functions which only exist at values of a non-continuous argument, most often integral values?

Is there any difference between a limit being infinity and not existing?

The number of ( in a problem should equal the number of ) which don't think they do at the moment.#1 Well not in the first line, I see it is OK further down.

Also, Mark44, doesn't that depend whether |α| ≥ 1 or not?

I think you can talk about limits of functions which only exist at values of a non-continuous argument, most often integral values?

Is there any difference between a limit being infinity and not existing?
well i mentioned domain of α...

Mark44
Mentor
Also, Mark44, doesn't that depend whether |α| ≥ 1 or not?
It says in post #1 that α ∈ (0, 1).

epenguin
Homework Helper
Gold Member Ooh Sorry, missed that. Well it looks like another problem like the OP's other one – product of a function tending to 0 and a tame finite function.

okay, I’m seriously confused about the presence of this oscillaing function.when do we say that limit does not tend to one fix position because its wildly oscillating?please help..

epenguin
Homework Helper
Gold Member
sin2(n!) is varying might as well be randomly as far as we are concerned, but again it remains in the range 0 to 1while whaT multiplies it gets ever smaller, it's the same thing as before. Looks like someone is trying to trap you by intervalling problems with unexpectedly easy ones and you are falling into it!

(I don't see this going to infinity for |α|<1).

Last edited:
• Mark44
Mark44
Mentor
Sin2(n!) is varying might as well be randomly as far as we are concerned, but again it remains in the range 0 to 1while whaT multiplies it gets ever smaller, it's the same thing as before. Looks like someone is trying to trap you by intervalling problems with unexpectedly easy ones and you are falling into it!
I agree completely.

okay, thanks to both of you