Limit of the form 0^infinity+infinity^0

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In summary, the limit of the form 0^infinity+infinity^0 is an indeterminate form that cannot be determined without additional context or information. This type of expression is known as an indeterminate form and cannot be evaluated using standard algebraic techniques. L'Hopital's rule cannot be used to determine the limit of this form, but other methods such as substitution, algebraic manipulation, and limit theorems can be applied. This limit has real-world applications in physics and engineering, particularly in thermodynamics and fluid dynamics, and can also be used to model certain functions as they approach infinity.
  • #1
abhineetK
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Homework Statement


What is the value of
lim {(1/x)sinx+(sinx)1/x} ,x>0 ?
x->0

Homework Equations


??

The Attempt at a Solution


lim {(1/x)sinx+(sinx)1/x}
x->0
=lim (1/x)sinx+lim (sinx)1/x
..x->0...x->0

Let A=lim (1/x)sinx
...x->0
and B=lim (sinx)1/x
...x->0

logA=lim (sinx)log(1/x)=0
...x->0
=>A=e0=1

logB=lim (1/x)log(sinx)=infinity
...x->0
=>B=infinity

The problem is that the answer is 0.
 
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  • #2
I understand that x goes to 0 in the limit, but how ? [itex] \nearrow 0 [/itex] or [itex] \searrow 0[/itex] ?
 
  • #3
Split it up into two limits and work with them. If both limits exist, then their sum will be the limit you want.

Since both expressions involve exponents, you are most likely going to have to take the log, which will probably get you something you can use L'Hopital's Rule on.
 
  • #4
I discussed this problem with someone..
He says,
lim (sinx)1/x=1
x->0
So, according to him answer must be 2.

I worked it out again and got the same answer, i.e., infinity...
How is it happening...??
Please, show the steps if you have got the answer 2, 1 or 0.
These were the options given to me...
However, if you feel that the correct answer is not among the options please tell with explanation and steps.
 
  • #5
I believe the limit in your first post is 1, but I can't justify it just yet. (sin x)^(1/x) -->0 as x -->0+, and (1/x)^(sin x) --> 1 as x --> 0+. I'll see if I can provide some justification for this result.
 
  • #6
bigubau said:
I understand that x goes to 0 in the limit, but how ? [itex] \nearrow 0 [/itex] or [itex] \searrow 0[/itex] ?
As x approaches 0 from above. (The OP said that x > 0.)
 
  • #7
Let y1 = (1/x)sin x
Then ln y1 = ln[(1/x)sin x] = sinx*ln(1/x) = sinx*(-lnx) = -lnx/cscx
In the next step I use L'Hopital's Rule, since we have an indeterminate form of [infinity/infinity]. Also, in the following steps, lim means limit as x -->0+
lim ln[ ln y1] = lim [-lnx/cscx]
= lim [(-1/x)/(-csc x * cot x)] = lim[(1/x)/(csc x * cot x)]
= lim[(1/x)/(1/sinx * cos x/sinx)] = lim [sin^2(x)/(x cos x)]
= lim [(sin(x)/x * tan(x)] = lim[sin(x)/x]*lim tan(x) = 1*0 = 0.

Since lim ln y1 = 0, ln lim y1 = 0, hence lim y1 = 1.

The other limit is easier.
Let y2 = sinx1/x = (eln(sinx))1/x = e(1/x)ln(sin x)

Before taking the limit, let's look at what the factors in the exponent on e are doing as x --> 0+.

1/x --> + infinity.
sin(x) --> 0+, so ln(sin(x)) --> -infinity.
The product of 1/x and ln(sin(x)) --> -infinity.
Since the exponent on e is approaching -infinity, then e to that power --> 0.
Hence lim y2 = lim e(1/x)ln(sin x) = 0.

Therefore lim (y1 + y2) = 1 + 0 = 1.
 

Related to Limit of the form 0^infinity+infinity^0

What is the limit of the form 0^infinity+infinity^0?

The limit of the form 0^infinity+infinity^0 is an indeterminate form. This means that it cannot be determined without further context or information.

What is an indeterminate form?

An indeterminate form is a mathematical expression that cannot be evaluated with standard algebraic techniques. It requires additional information or context to determine its value.

Can the limit of the form 0^infinity+infinity^0 be determined using L'Hopital's rule?

No, L'Hopital's rule can only be applied to limits of the form 0/0 or infinity/infinity. It cannot be used to determine the limit of the form 0^infinity+infinity^0.

What other methods can be used to determine the limit of the form 0^infinity+infinity^0?

Other methods that can be used to determine the limit of the form 0^infinity+infinity^0 include substitution, algebraic manipulation, and the use of other limit theorems such as the squeeze theorem or the Cauchy condensation test.

Are there any real-world applications of the limit of the form 0^infinity+infinity^0?

Yes, this limit often arises in physics and engineering calculations, particularly in problems related to thermodynamics and fluid dynamics. It can also be used to model the behavior of certain functions, such as the gamma function, as they approach infinity.

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