How to calculate this limit?

Mar 2010
26
1
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.

thanks in advance...
 

Attachments

Jan 2010
354
173
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
Apr 2005
20,249
7,909
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.
 
Oct 2009
4,261
1,836
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.

thanks in advance...

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

Tonio
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
\(\displaystyle \ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}\) , 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
Dec 2007
16,948
6,768
Zeitgeist
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.
 
Oct 2009
4,261
1,836
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​
 
Mar 2010
26
1
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 :

\(\displaystyle \ln(1+\sin x-x)^{1/x^3}=\frac{\ln(1+\sin x-x)}{x^3}
\)
 
Oct 2009
4,261
1,836
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
 
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Jan 2010
354
173
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.