Limit of 0^0: Evaluating x^sinx

[Krushnaraj Pandya]

Homework Statement

lim x--->0 |x|^sinx is?

Homework Equations

lim x-->0 f(x)^g(x), if both functions tend to 0, limit is equal to e^log[f(x).g(x)] with the same limit..(i)

The Attempt at a Solution

when x>0, it is x^sinx and x<0 it is -1/x^sinx. putting the first case in (i) we get e^log(x.sinx), since log0 is minus infinite, the answer should be 0 but its 1, I don't know how to evaluate the second case either. I'd appreciate some help, thank you

 

[andrewkirk]

Using that equation just makes more work. Instead, take it back to first principles and use an epsilon-delta argument. Show that, for any epsilon>0 we can find a delta>0 such that if $|x-0|<$delta, we will have $|x|\sin x<$epsilon.

That way is easy.

 

[WWGD]

I think your approach could be productive if you use the approximation sinx ~x when x is close to 0. Then you get $e^{xlnx}$ . Then maybe try L'Hopital after working with the exponents.

 

[Krushnaraj Pandya]

Using that equation just makes more work. Instead, take it back to first principles and use an epsilon-delta argument. Show that, for any epsilon>0 we can find a delta>0 such that if $|x-0|<$delta, we will have $|x|\sin x<$epsilon.

That way is easy.

I was trying to find why my approach is wrong. What makes it harder is that this isn't in my school syllabus and is something I'm studying additionally for competitive exams so I've just memorized some equations for some forms and some popular techniques and taylor expansions. Does this equation not apply universally?

 

[WWGD]

I suggest you don't look for things to apply universally. There are too many weird turns in Math. But here, when x is small , you have $Sinx$ ~ $x$. Then your expression becomes close to $x^x$, and remember $0^0=1$.

 

[andrewkirk]

Does this equation not apply universally?

I have never seen that rule and it doesn't look correct to me. Perhaps it is not written correctly above? Indeed, whct is the difference between e^log[f(x).g(x)] and f(x).g(x). It seems to me they say exactly the same thing, except that there are sections of the real numbers where the former does not even exist.

If you want to know why the rule doesn't work, the first step would be to find the source where you got it from post write the rule here, very carefully and verbatim.

 

[Krushnaraj Pandya]

I think your approach could be productive if you use the approximation sinx ~x when x is close to 0. Then you get $e^{xlnx}$ . Then maybe try L'Hopital after working with the exponents.

I don't understand how you got e^xlnx, what I tried was dividing and multiplying this by e^x- this gave form e^(xlnx+1)/e^x xlnx is 0 when lim x->0 this gives another wrong answer e

 

[WWGD]

I don't understand how you got e^xlnx, what I tried was dividing and multiplying this by e^x- this gave form e^(xlnx+1)/e^x xlnx is 0 when lim x->0 this gives another wrong answer e

I am using the approximation sinx~x when x is small. You can remember if you know the limit $(Sinx/x)$ ~1 as x->0.. You can also try the Taylor series and see that the terms beyond the first go to 0 quickly.

 

[Krushnaraj Pandya]

I am using the approximation sinx~x when x is small. You can remember if you know the limit $Sinx/x$ ~1 as x->0.. You can also try the Taylor series and see that the terms beyond the first go to 0 quickly.

I understand sinx/x tends to 1, what I don't understand is how that approximates to x^x

 

[Krushnaraj Pandya]

I have never seen that rule and it doesn't look correct to me. Perhaps it is not written correctly above? Indeed, whct is the difference between e^log[f(x).g(x)] and f(x).g(x). It seems to me they say exactly the same thing, except that there are sections of the real numbers where the former does not even exist.

If you want to know why the rule doesn't work, the first step would be to find the source where you got it from post write the rule here, very carefully and verbatim.

Look at (ii) under remarks

 

[WWGD]

Let me go to sleep, it is kind of late. Will get back tmw.

 

[Krushnaraj Pandya]

Let me go to sleep, it is kind of late. Will get back tmw.

No problem! I just woke up here, weird how people help everyone across so much distance everyday. PF awes me sometimes. Thank you for your help :D

 

[Krushnaraj Pandya]

whct is the difference between e^log[f(x).g(x)] and f(x).g(x)

We are given f(x)^g(x), not f(x).g(x) so its not exactly the same...but it does seem counter-intuitive

 

[andrewkirk]

Look at (ii) under remarks

Ah, from that image I see that I missed your '^' sign in the OP, so I thought you were multiplying. THat's the trouble with not using LaTeX in presenting problems.

With that info on board, my observation is that you have misplaced brackets. The item you want to work with is $e^{(\log x)\cdot (\sin x)}$, not $e^{\log (x\cdot \sin x)}$, which is what you wrote.

We can use the fact that $\lim_{x\to 0} x\cdot \log x=0$, together with the small angle approximation $x\approx \sin x$ that WWGD suggests, to show that the exponent goes to zero, so that the expression goes to 1.

 

[Krushnaraj Pandya]

Oh, I just realized I overlooked a bracket in the equation...really sorry. Let me try again and get back. I see you just posted that, thank you

 

[Krushnaraj Pandya]

Ok, so I got 1 this way. Now considering the case when x<0, $f(x) = \frac{-1}{x^{sinx}}$ We just determined that x^sinx tends to 1 when x tends to 0, so f(x)= -1, does this mean the limit doesn't exist?

 

[WWGD]

Sorry, I am back for s bit .idea is , if sinx~x , then. x^{sinx}~x^x =e^{xlnx} rewrite exponent so you can use L'Hopital asnd find overall answer.

 

[Krushnaraj Pandya]

Sorry, I am back for s bit .idea is , if sinx~x , then. x^{sinx}~x^x =e^{xlnx} rewrite exponent so you can use L'Hopital asnd find overall answer.

That's what I did...and got the correct answer as well. Thank you :D now I'm thinking about the case when x<0, where it gives -1 (see post no. 16)

 

[Delta2]

Ok, so I got 1 this way. Now considering the case when x<0, $f(x) = \frac{-1}{x^{sinx}}$ We just determined that x^sinx tends to 1 when x tends to 0, so f(x)= -1, does this mean the limit doesn't exist?

You go wrong here, when x<0 it is $f(x)=(-x)^{\sin x}=(-x)^{-\sin(-x)}=\frac{1}{(-x)^{\sin(-x)}}$ and the limit is again 1 cause $(-x)^{\sin(-x)}$ for x<0 is the same as $x^{\sin x}$ for x>0 (change of variable y=-x).

 

[Krushnaraj Pandya]

You go wrong here, when x<0 it is $f(x)=(-x)^{\sin x}=(-x)^{-\sin(-x)}=\frac{1}{(-x)^{\sin(-x)}}$ and the limit is again 1 cause $(-x)^{\sin(-x)}$ for x<0 is the same as $x^{\sin x}$ for x>0 (change of variable y=-x).

I made very silly mistakes throughout this entire thread, thank you everyone for your patience. I just hope I don't do that in the exam...