# How to calculate this limit?

#### azarue

Hi all forum members, long time...

I've been struggling with a limit problem that involves L'Hôpital's rule...

is it something that I can solve using this assumption ?

the limit in the image is equal to lim(x->0) e^(ln(1+sinx-x))^(1/(x^3))??

please check the attached file.

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#### drumist

The goal is to manipulate the expression so that we get a $$\displaystyle \frac{0}{0}$$ or $$\displaystyle \frac{\infty}{\infty}$$ condition, which is necessary to use L'Hopital's rule. Start by setting the expression equal to some variable $$\displaystyle A$$ (so that we remember what we are solving for) and then do a few algebraic manipulations:

$$\displaystyle A = \lim_{x \to 0} \left[ (1+\sin x - x)^{1/x^3} \right]$$

$$\displaystyle \ln A = \ln \left( \lim_{x \to 0} \left[ (1+\sin x - x)^{1/x^3} \right] \right)$$

$$\displaystyle \ln A = \lim_{x \to 0} \left( \ln \left[ (1+\sin x - x)^{1/x^3} \right] \right)$$

$$\displaystyle \ln A = \lim_{x \to 0} \left[ \frac{1}{x^3} \ln(1+\sin x - x) \right]$$

$$\displaystyle \ln A = \lim_{x \to 0} \left[ \frac{\ln(1+\sin x - x)}{x^3} \right]$$

So from this point, we can use L'Hopital's to solve the right-hand side of the above equation. Notice that we're actually solving for $$\displaystyle \ln A$$ so you'll have one final step to get back to $$\displaystyle A$$ which is the solution.

#### HallsofIvy

MHF Helper
The goal is to manipulate the expression so that we get a $$\displaystyle \frac{0}{0}$$ or $$\displaystyle \frac{\infty}{\infty}$$ condition, which is necessary to use L'Hopital's rule. Start by setting the expression equal to some variable $$\displaystyle A$$ (so that we remember what we are solving for) and then do a few algebraic manipulations:

$$\displaystyle A = \lim_{x \to 0} \left[ (1+\sin x - x)^{1/x^3} \right]$$
Sorry, but this won't work. The limit is not
$$\displaystyle \lim_{x\to 0}e^{ln((1+ sinx- x)^{1/x^3})}$$

It is
$$\displaystyle \lim_{x\to 0}e^{(ln(1+sin x-x))^{1/x^3}}$$
That is, the entire ln expression is to the $$\displaystyle 1/x^3$$ power.

$$\displaystyle \ln A = \ln \left( \lim_{x \to 0} \left[ (1+\sin x - x)^{1/x^3} \right] \right)$$

$$\displaystyle \ln A = \lim_{x \to 0} \left( \ln \left[ (1+\sin x - x)^{1/x^3} \right] \right)$$

$$\displaystyle \ln A = \lim_{x \to 0} \left[ \frac{1}{x^3} \ln(1+\sin x - x) \right]$$

$$\displaystyle \ln A = \lim_{x \to 0} \left[ \frac{\ln(1+\sin x - x)}{x^3} \right]$$

So from this point, we can use L'Hopital's to solve the right-hand side of the above equation. Notice that we're actually solving for $$\displaystyle \ln A$$ so you'll have one final step to get back to $$\displaystyle A$$ which is the solution.

#### tonio

Hi all forum members, long time...

I've been struggling with a limit problem that involves L'Hôpital's rule...

is it something that I can solve using this assumption ?

the limit in the image is equal to lim(x->0) e^(ln(1+sinx-x))^(1/(x^3))??

please check the attached file.

$$\displaystyle \ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}$$ , and now do L'Hospital with this thing!(Wink)

Tonio

#### HallsofIvy

MHF Helper
$$\displaystyle \ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}$$ , and now do L'Hospital with this thing!(Wink)

Tonio
That is what is shown in the attachment to the original post, but in the post itself it is $$\displaystyle (ln(1+ sin x- x))^{1/x^3}$$

#### mr fantastic

MHF Hall of Fame
That is what is shown in the attachment to the original post, but in the post itself it is $$\displaystyle (ln(1+ sin x- x))^{1/x^3}$$
The limit has the form $$\displaystyle \lim_{x \to 0} f(x)^{g(x)} = \lim_{x \to 0} e^{\ln f(x)^{g(x)}} = \lim_{x \to 0} e^{g(x) \ln f(x)}$$ so I think what's been posted is OK.

#### tonio

Sorry, but this won't work. The limit is not
$$\displaystyle \lim_{x\to 0}e^{ln((1+ sinx- x)^{1/x^3})}$$

It is
$$\displaystyle \lim_{x\to 0}e^{(ln(1+sin x-x))^{1/x^3}}$$
That is, the entire ln expression is to the $$\displaystyle 1/x^3$$ power.

I think you misread: $$\displaystyle a^x=e^{x\ln a}=e^{\ln a^x}$$ ... a power x applies only to the argument a of the logarithm and not to the whole logarithm.​

Tonio​

#### azarue

Thanks for all of your responses.

i got a few answers and I'm a bit confused.

would it be fine to go ahead and start working with this expression :

$$\displaystyle \lim_{x\to 0}e^{ln((1+ sinx- x)^{1/x^3})}$$

or should I take the other opinion and work with the expression :

$$\displaystyle \ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}$$

#### tonio

i got a few answers and I'm a bit confused.

would it be fine to go ahead and start working with this expression :

$$\displaystyle \lim_{x\to 0}e^{ln((1+ sinx- x)^{1/x^3})}$$$$\displaystyle$$

If you want...(Giggle)...but I can't see how you'll manage to do something with this expression.

or should I take the other opinion and work with the expression :

$$\displaystyle \ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}$$$$\displaystyle$$

Don't you think this way is much easier?

Tonio9

azarue

#### drumist

That is what is shown in the attachment to the original post, but in the post itself it is $$\displaystyle (ln(1+ sin x- x))^{1/x^3}$$
I believe that what he wrote out was in error and that the expression in the image was the intended problem. I think he meant to write was

$$\displaystyle \lim_{x \to 0} e^{\ln \left[ (1+\sin x - x)^{1/x^3} \right] }$$

but he put the parentheses in the wrong spot.