# Math Help - How to calculate this limit?

1. ## How to calculate this limit?

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))??

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

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

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

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

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

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

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 $\ln A$ so you'll have one final step to get back to $A$ which is the solution.

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

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

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

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

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

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

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

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 $\ln A$ so you'll have one final step to get back to $A$ which is the solution.

4. Originally Posted by 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))??

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

Tonio

5. Originally Posted by tonio
$\ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}$ , and now do L'Hospital with this thing!

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

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

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

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

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

Tonio

8. ## 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 :

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

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

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

9. Originally Posted by azarue
i got a few answers and I'm a bit confused.

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

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

If you want......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 :

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

Don't you think this way is much easier?

Tonio9

10. Originally Posted by HallsofIvy
That is what is shown in the attachment to the original post, but in the post itself it is $(ln(1+ sin x- x))^{1/x^3}$
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

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

but he put the parentheses in the wrong spot.

11. Originally Posted by azarue
i got a few answers and I'm a bit confused.

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

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

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

$\ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}
$
If you prefer to do it this way, it is fine. The reason I did it differently is that I prefer to not have limits in the exponent. For one thing it gets hard to write!

This would be more or less the process if you are doing it like that:

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

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

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

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

You can see the markup is hard to understand and small. If I instead use $\exp(x) = e^x$, then we can write it a little more clearly:

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

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

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

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

This is a little easier to understand at least. Does this help?

12. ## Yes

Guys, I might had a type while writing the expression. please refer to the attached file. this is the correct exercise. sorry...

Tonio, I decided to go ahead and work with both of them, and no doubt about it, you right it is easier. thanks again

I got to this limit :

$\lim_{x\to0}\left(\sin x-x+1\right)^{{{1}\over{x^3}}} = e^ {- {{1}\over{6}} }$

It seems fine, I will not post the whole solution because I'm still new to the math typing..