#Limit should be 1 but geogebra shows it's around 0.8 I think
wanton falcon
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crimson birch
what have you tried?
wanton falcon
The limit above the fraction looks exactly like e^n
And dividing with the e^n below the fraction ought to give the limit of 1 so it is 1
But I am not sure if this reasoning is correct
Because geobra kinda shows it doesn't go above the line y = 0.8 so I am having doubts
crimson birch
And dividing with the e^n below the fraction ought to give the limit of 1 so it ...
its not 1, just to let you know
wanton falcon
Wtf
wanton falcon
Please do
crimson birch
,w Limit[Divide[Power[(40)Divide[2m-1,2m-2](41),2Power[m,2]-2m],Power[e,m]],m->infinity]
cosmic monolithBOT
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cosmic monolithBOT
dark matter
wooden monolith
To be clear, this is $\lim_{n \to \infty} \frac{\left(\frac{2n - 1}{2n - 2}\right)^{2n^2 - 2n}}{e^n}$? (I really hope I typed that correctly.)
cosmic monolithBOT
Techie Literate
wooden monolith
So I immediately notice the numerator is a 1^infinity indeterminate form.
wooden monolith
So my first instinct is to evaluate just that limit.
Because if it's finite, then the whole limit is just 0, because the denominator goes to infinity.
And if it's infinite, that opens up L'Hopital's rule for use.
wanton falcon
It is infinite since it goes to e^n in the limit which diverges
wooden monolith
Can you prove that?
wooden monolith
I'm not wholly convinced that logic holds, since you're evaluating one part of the limit before another part.
You should use the standard method of evaluating a 1^infinity indeterminate form.
wooden monolith
I keep getting 3 different answers
Show your work.
wooden monolith
So it is e^(n * ln( (1+1/2n-2)^(2n-2) ) )
...why don't you take the whole exponent out into the coefficient of the logarithm?
wanton falcon
e^(2n(n-1) * ln(1+1/(2n-2)) )
But this is a infinity * 0 situation
So I thought of transforming the ln(...) into 1/(1/ln(...)) so that we are now into a inf/inf situation
And now we could use L'Hopital?
wooden monolith
Sure, but usually you want to double-reciprocal the polynomial, not the logarithm.
Because d/dx(1/ln(x)) contains an (ln(x))^2 term.
wanton falcon
Ah yes you're right
wooden monolith
That way you wind up with just a rational function.
wooden monolith
I'm not sure I follow all of your cancellations, but I also got a cubic over a quadratic.
wanton falcon
Exactly
But Wolframalpha says it converges to -1/4 somehow. So does Geogebra indicate so
I'll sleep on it
Thanks for your help so far anyway
wooden monolith
Actually, your whole error here is improperly evaluating a 1^inf indeterminate form.
Because the original limit can be rearranged to $\lim_{n \to \infty}\left(\frac{\left(\frac{2n - 1}{2n - 2}\right)^{2n - 2}}{e}\right)^n$
cosmic monolithBOT
Techie Literate
wooden monolith
Because the original limit can be rearranged to $\lim_{n \to \infty}\left(\frac{...
@wanton falcon
wanton falcon
ye
wooden monolith
So in general you can't partially evaluate a limit.
wanton falcon
But you can do that when the parts are both convergent, right? That has to be true at least
wooden monolith
No, because 0^0 is also indeterminate.
As is 0/0.
But obviously, a limit that converges to 0... converges.
wanton falcon
I see, thank you
wooden monolith
I see, thank you
Just for future reference, the rule is that the limit of a sum/product of functions equals the sum/product of the limits of the functions if the limits of both functions converge. Every indeterminate form ultimately amounts to a sum/product of functions whose limits do not both converge.
scenic sandal
Let me try to rewrite the whole thing in one place
\begin{align*}
\ln \left( \frac{\left( \frac{2n-1}{2n-2}\right)^{2n^2-2n}}{e^n}\right) &=
(2n^2 - 2n) \ln \left( 1 +\frac{1}{2n-2} \right) - n \\
&= n \left((2n-2) \ln \left(1 + \frac{1}{2n-2} \right) - 1 \right)
\end{align*}
Nagi
scenic sandal
I think you can finish this with l'Hospital, as you've mentioned
\begin{align*}
n \left((2n-2) \ln \left(1 + \frac{1}{2n-2} \right) - 1 \right) &= \frac{n}{2n-2} \times \frac{(2n-2) \ln \left(1 + \frac{1}{2n-2} \right) - 1}{\frac{1}{2n-2}}
\end{align*}
Nagi
scenic sandal
I'll leave it to you to finish the computations here
wooden monolith
OP's asleep.
scenic sandal
But Wolframalpha says it converges to -1/4 somehow. So does Geogebra indicate so
I also find this after finishing the computations on my side, just for documentation purposes
wanton falcon
😴
scenic sandal
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@wanton falcon
wanton falcon
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