Limit of the form 0^infinity+infinity^0

[abhineetK]

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.

 

[dextercioby]

I understand that x goes to 0 in the limit, but how ? \nearrow 0 or \searrow 0 ?

 

[Mark44]

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.

 

[abhineetK]

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.

 

[Mark44]

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.

 

[Mark44]

I understand that x goes to 0 in the limit, but how ? \nearrow 0 or \searrow 0 ?

As x approaches 0 from above. (The OP said that x > 0.)

 

[Mark44]

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 = sinx^1/x = (e^ln(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.