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

In summary: function so it would depend on the definition of "exist" but in general, if a limit does not exist then it can't be found by any function that you might know of
  • #1
randomgamernerd
139
4

Homework Statement

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

Homework Equations

: -[/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 please.
 
Last edited:
Physics news on Phys.org
  • #2
randomgamernerd said:

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

Homework Equations

: -[/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 please.
 
  • #3
i
Mark44 said:
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
 
  • #4
I think I have corrected all the typos...please help me out now @Mark44
 
  • #5
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.
 
  • #6
Mark44 said:
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
 
  • #7
Mark44 said:
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?
 
  • #8
epenguin said:
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...
 
  • #9
epenguin said:
Also, Mark44, doesn't that depend whether |α| ≥ 1 or not?
It says in post #1 that α ∈ (0, 1).
 
  • #10
:redface: 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.
 
  • #11
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..
 
  • #12
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:
  • Like
Likes Mark44
  • #13
epenguin said:
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.
 
  • #14
okay, thanks to both of you
 

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

1. What is the definition of a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as the input approaches a certain value. It represents the value that the function approaches, or "goes towards," as the input gets closer and closer to the specified value.

2. How do you find the limit of a function as the input approaches infinity?

To find the limit of a function as the input approaches infinity, you can use the concept of "infinity" as a very large number. You can then substitute this very large number in place of the input and evaluate the function to see what value it approaches.

3. What is the significance of x^α.sin²(x)/(x+1) in the limit as x approaches infinity?

The function x^α.sin²(x)/(x+1) represents the behavior of a function as the input (x) approaches infinity. It is an important function in calculus and is often used to model real-world phenomena such as population growth and decay.

4. What does it mean when the limit of a function as x approaches infinity is undefined?

When the limit of a function as x approaches infinity is undefined, it means that the function does not approach a specific value as the input gets larger and larger. This could be due to various factors such as a vertical asymptote or a non-existent limit.

5. How can you use L'Hopital's Rule to find the limit of x^α.sin²(x)/(x+1) as x approaches infinity?

L'Hopital's Rule is a useful mathematical tool for finding the limit of a function as the input approaches infinity. It states that if the limit of the ratio of two functions is indeterminate (i.e. gives a value of "0/0" or "∞/∞"), then the limit of the original function can be found by taking the limit of the derivatives of the two functions. In the case of x^α.sin²(x)/(x+1), this would involve taking the limit of the derivatives of x^α and sin²(x) and then dividing the two values.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
795
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
5K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
886
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top