L'Hopital Limit with rational exponent

[crybllrd]

Homework Statement

I need to find the limit of this using L'Hopital Rule, if it exists.

$lim_{x\rightarrow0^{+}}(\frac{sinx}{x})^{1/x^{2}}$

Homework Equations

All of our examples used lim f(x)/g(x)=f'(x)/g'(x)

The Attempt at a Solution

We have not dealt with any limits like this that have a rational exponent, and I'm really not sure how to handle it.

Can someone please give me a theorem, because I have scoured the interwebs trying to find a similar example.

 

[kmikias]

why can't you rewrite the the limit. example, 1/x can be rewritten as x^-1. another example (sin(x) / x )^x can be rewritten as sin(x)^x / x^x.

After that, you can use L'Hopital rule which you take the derivative of the top and the derivative of the bottom then you will plug the limit value and again if you get 0 in the bottom ...you use L'Hopital rule until you avoid 0 in the denominator .

 

[crybllrd]

Ha, that's perfect.
I didn't even think about it.
Thanks a lot!

 

[Curious3141]

Hint: use the Taylor's series for sin x.

EDIT: Using the Taylor's series enables a quick solution without the need for LH Rule. Of course, if the question stipulated that you need to use LH Rule, then you've no choice.

 

[crybllrd]

Uh oh, this doesn't seem to be right:$$lim_{x\rightarrow0^{+}}(\frac{sinx}{x})^{1/x^{2}}$$$$lim_{x\rightarrow0^{+}}(\frac{sinx}{x^{3}})=0/0$$[tex]lim_{x\rightarrow0^{+}}(\frac{cosx}{3x$^{2}$}})=1/0[/tex]$$lim_{x\rightarrow0^{+}}(\frac{-sinx}{6x})=0/0$$$$lim_{x\rightarrow0^{+}}(\frac{-cosx}{6}})=-1/6$$

To me it looks correct, but I didn't use a taylor Series like Curious suggested. And my answer doesn't make use of the x->0+ approach from the right.
Did I miss something?

EDIT: LaTeX is goofing up, my final answer was -1/6

 

[gb7nash]

$$lim_{x\rightarrow0^{+}}(\frac{sinx}{x^{3}})$$

How did you get this?

In any case, what makes this problem more complex is the 1/x² exponent, so it's not as simple as just applying l'hopitals rule. I would take the natural log in the beginning, find the limit, and apply exponentiation.

 

[crybllrd]

...take the natural log in the beginning, find the limit, and apply exponentiation.

I realize lim x->0+ needs to be in every step, but for the sake of my sanity I will leave them off in this problem (of course I will write it down on my homework though)

$\frac{1}{x^{2}}ln\frac{sinx}{x}=\frac{ln\frac{sinx}{x}}{x^{2}}=\frac{lnsinx-lnx}{x^{2}}=indeterminate$$\frac{\frac{1}{sinx}cosx-\frac{1}{x}}{2x}=\frac{cotx-\frac{1}{x}}{2x}$At this point, the fraction in the numerator's x will raise a power every derivative, always resulting in a divide by zero.
Not sure where I messed up.

 

[Curious3141]

OK, crybllrd, since you've shown effort (and quite a lot of it! ), I'll show you how I did it. I used Taylor's series repeatedly without any LH rule.

The limit to be worked out is ${(\frac{\sin x}{x})}^{\frac{1}{x^2}}$ as x tends to 0 from the right.

Taylor series: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} -...$

So the expression becomes:

${(1 - \frac{x^2}{3!} + \frac{x^4}{5!} -...)}^{\frac{1}{x^2}} = {[1 +x^2(-\frac{1}{3!} + \frac{x^2}{5!} - \frac{x^4}{7!} +...)]}^{\frac{1}{x^2}}$

Let this mess: $(-\frac{1}{3!} + \frac{x^2}{5!} - \frac{x^4}{7!} +...)$ be denoted $y$.

Applying Binomial Theorem (which is also a variant of the relevant Taylor series), the expression becomes:

$1 + \frac{1}{x^2}.x^2.y + \frac{(\frac{1}{x^2})(\frac{1}{x^2}-1)}{2!}.y^2.x^4 + ...$

You'll notice that all expressions of the form (1/x^2)(1/x^2 - 1)...(1/x^2 - k).(x^2(k+1)) will reduce to 1 at the limit x -> 0.

That just leaves:

$1 + y + \frac{y^2}{2!} + ...+\frac{y^n}{n!}+... = e^y$

Now limit of y as x -> 0 = -1/3! = -1/6.

So the final answer is $e^{-\frac{1}{6}}$

 

[SammyS]

I realize lim x->0+ needs to be in every step, but for the sake of my sanity I will leave them off in this problem (of course I will write it down on my homework though)

$\displaystyle\frac{1}{x^{2}}ln\frac{sinx}{x}=\frac{ln\frac{sinx}{x}}{x^{2}}=\frac{lnsinx-lnx}{x^{2}}=indeterminate$

$\displaystyle\frac{\frac{1}{sinx}cosx-\frac{1}{x}}{2x}=\frac{cotx-\frac{1}{x}}{2x}$

At this point, the fraction in the numerator's x will raise a power every derivative, always resulting in a divide by zero.
Not sure where I messed up.

\displaystyle
Rewrite $\displaystyle\frac{cotx-\frac{1}{x}}{2x}$ as:

$\displaystyle\frac{x\frac{\cos(x)}{sin(x)}-1}{2x^2}=\frac{x\cos(x)-\sin(x)}{2x^2\sin(x)}$

A couple more applications of L'Hôpital's rule should do the trick !

 

[crybllrd]

Thanks a lot guys.
My teacher did say that we must use L'Hopital's rule to do it, but I got it worked out with a final answer of e^(-1/6) after exponentiation.

Thanks again!

 

[Curious3141]

Thanks a lot guys.
My teacher did say that we must use L'Hopital's rule to do it, but I got it worked out with a final answer of e^(-1/6) after exponentiation.

Thanks again!

As I said, if you were instructed to use LHR here, you had little choice. But I find that LHR, while useful in many cases, is sometimes not the method of choice. In this problem, for instance, you need to rearrange the quotients just right so that they become simpler with every application of LHR. And LHR often obscures the manner in which the limit is arrived at. With so many manipulations and repeated applications, I bet you can't tell exactly "how" you got to that (-1/6) before exponentiating. In contrast, with the more direct Taylor series method, you can pretty much "see" exactly what terms are vanishing at every stage, it's just algebra.

In any case, you now have two perfectly good methods to tackle problems like this, and it's always good to have a choice!