# Limit using L'Hôpital

#### Ulysses

Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}$$

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}$$

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}$$

And then:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}$$

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}$$

From here it would seem to be that the answer is $$\displaystyle e^{-\infty}$$ (which I think it would be equal to zero) but I used derive to make the calculus, and the límit should give $$\displaystyle e^{-1/6}$$

Bye there.

#### tonio

Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}$$

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}$$

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}$$

And then:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}$$

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}$$

Bye there.

$$\displaystyle \left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2}$$ $$\displaystyle =\frac{x\cos x-\sin x}{x\sin x}$$ , so applying L'Hospital you get:

$$\displaystyle \frac{x\cos x-\sin x}{2x^2\sin x}$$ , and L'H once again: $$\displaystyle \frac{-x\sin x}{4x\sin x+2x^2\cos x}$$ , and once again: $$\displaystyle \frac{-\sin x-x\cos x}{4\sin x+8x\cos x-2x^2\sin x}$$ , and again:

$$\displaystyle \frac{-2\cos x+x\sin x}{12\cos x-12x\sin x-2x^2\cos x}\xrightarrow [x\to o^+]{}-\frac{2}{12}=-\frac{1}{6}$$ , so the limit is $$\displaystyle e^{-1/6}$$

Tonio

• Ulysses

#### Ulysses

Thank you very much Tonio!

#### Prove It

MHF Helper
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}$$

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}$$

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}$$

And then:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}$$

How should I proceed?

Bye there.
$$\displaystyle \lim_{x \to 0^{+}}\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\right] = \lim_{x \to 0^{+}}e^{\ln{\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\right]}}$$

$$\displaystyle = \lim_{x \to 0^{+}}e^{\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}}$$

$$\displaystyle = e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}}$$

This $$\displaystyle \to \frac{0}{0}$$, so you can use L'Hospital's Rule...

$$\displaystyle e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}} = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{\frac{x\cos{x} - \sin{x}}{x^2}}{\frac{\sin{x}}{x}}}{2x}\right)}$$

$$\displaystyle = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{x\cos{x} - \sin{x}}{x\sin{x}}}{2x}\right)}$$

$$\displaystyle = e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)}$$.

This still $$\displaystyle \to \frac{0}{0}$$ so use L'Hospital's Rule again...

$$\displaystyle e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} + 4x\sin{x}}\right)}$$

$$\displaystyle = e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)}$$

This still $$\displaystyle \to \frac{0}{0}$$, so use L'Hospital's Rule again...

$$\displaystyle e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(-\frac{\cos{x}}{4\cos{x} + 2x\sin{x} + 2\cos{x}}\right)}$$

$$\displaystyle = e^{-\frac{1}{4 + 2}}$$

$$\displaystyle = e^{-\frac{1}{6}}$$.

PHEW!

Edit: Fixed a small mistake.

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#### Prove It

MHF Helper
$$\displaystyle \left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2}$$ $$\displaystyle =\frac{x\cos x-\sin x}{x\sin x}$$ , so applying L'Hospital you get:

$$\displaystyle \frac{x\cos x-\sin x}{2x^2\sin x}$$ , and L'H once again: $$\displaystyle \frac{-x\sin x}{4x\sin x+2x^2\cos x}$$ , and once again: $$\displaystyle \frac{-\sin x-x\cos x}{4\sin x+8x\cos x-2x^2\sin x}$$ , and again:

$$\displaystyle \frac{-2\cos x+x\sin x}{12\cos x-12x\sin x-2x^2\cos x}\xrightarrow [x\to o^+]{}-\frac{2}{12}=-\frac{1}{6}$$ , so the limit is $$\displaystyle e^{-1/6}$$

Tonio
You've forgotten about the $$\displaystyle 2x$$ in the denominator that comes from the power in the logarithm.

#### chiph588@

MHF Hall of Honor
In general to get the $$\displaystyle 1^\infty$$ indeterminate form into the form $$\displaystyle \frac00$$ do the following:

If $$\displaystyle \lim_{x\to a}f(x) = 1$$ and $$\displaystyle \lim_{x\to a}g(x) = \infty$$

then $$\displaystyle \lim_{x\to a} f(x)^{g(x)} = \exp\left(\lim_{x\to a}\frac{\ln f(x)}{1/g(x)}\right)$$.

#### Ulysses

I wanted to ask him about that Prove it. Thanks. And I know that the result that he get to its good, because its the same that Derive gave it to me.

I have to go now, but I'll read this later.

Bye there, and thanks to both of you.

#### simplependulum

MHF Hall of Honor
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}$$

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}$$

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}$$

And then:

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}$$

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

$$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}$$

From here it would seem to be that the answer is $$\displaystyle e^{-\infty}$$ (which I think it would be equal to zero) but I used derive to make the calculus, and the límit should give $$\displaystyle e^{-1/6}$$

Bye there.

We have $$\displaystyle \frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...$$

$$\displaystyle \lim_{x\to 0} \left( \frac{\sin(x)}{x} \right)^{\frac{1}{x^2}} = \lim_{x\to 0} \left( 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...\right)^{\frac{1}{x^2}}$$

Let $$\displaystyle n = \frac{1}{x^2}$$ and we have

$$\displaystyle \lim_{n \to \infty} \left( 1 - \frac{1}{n}( \frac{1}{6} - \frac{1}{120 n} - ...) \right )^n$$

$$\displaystyle = exp \{\ -( \lim_{n \to \infty} \frac{1}{6} - \frac{1}{120 n} - ... ) \}\$$

$$\displaystyle = e^{-1/6}$$

EDIT: I think it is unnecessary to evaluate only the right hand limit .

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#### simplependulum

MHF Hall of Honor
$$\displaystyle e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} - 4x\sin{x}}\right)}$$
Mistake appears in this line

The derivative of $$\displaystyle 2x^2 \sin(x)$$ is

$$\displaystyle 4x\sin(x) + 2x^2 \cos(x)$$

#### Prove It

MHF Helper
Mistake appears in this line

The derivative of $$\displaystyle 2x^2 \sin(x)$$ is

$$\displaystyle 4x\sin(x) + 2x^2 \cos(x)$$
Thanks.