1. ## Interesting limit problem

We need to show the following,

Lim X-> 1

[X(X-1)T+1]^(1/(1-X)) = EXP(-T)

Obviously we need to apply L'Hopitals rule, but it is not straight forward how to proceed !

2. ## Re: Interesting limit problem

I would start by letting $y=[x(x-1)T+1]^{\frac{1}{1-x}}$ and then take the log of both sides and apply the property of logarithms to write it like this:

$ln(y)=\frac{1}{1-x}\ln[x(x-1)T+1]$

And note that you can take the limit of both sides and solve for the limit of $y$ as $x->1$

EDIT, typo:

$\lim_{x->1}ln(y)=\lim_{x->1}\frac{\ln[x(x-1)T+1]}{1-x}$

3. ## Re: Interesting limit problem

Yes, working from that starting point now.

4. ## Re: Interesting limit problem

Originally Posted by frustrated
We need to show the following,

Lim X-> 1

[X(X-1)T+1]^(1/(1-X)) = EXP(-T)
I simply have no idea what you mean. Have no idea what the function is.

5. ## Re: Interesting limit problem

Originally Posted by frustrated
Yes, working from that starting point now.
ok, note that I had to edit my post. Now, you know how to deal with the right hand side right? You mentioned L-hopitals rule which sounds right to me since it has the forum $\frac{0}{0}$ at $x=1$ so try that, and you can solve the equation for $\lim_{x->1}y$.

6. ## Re: Interesting limit problem

Originally Posted by Plato
I simply have no idea what you mean. Have no idea what the function is.

He's trying to show that the limit is equal to $e^{-T}$

7. ## Re: Interesting limit problem

Got it,

Rewrite as Ln y= Ln[x(x-1)T+1] / (1-x) then apply l'hopitals,

Rewrite Ln[x(x-1)T+1]=a (1-x)=b

Lim X->1 =da/dx / db/dx

da/dx = T(2x-1)/[Tx(x-1)+1]

db/dx = -1

da/dx / db/dx = -T(2x-1)/[Tx(x-1)+1] = -T

ln(y) = -T

Y= e^-T

Sorry for many edits !

8. ## Re: Interesting limit problem

Originally Posted by frustrated
Got it,

Rewrite as Ln y= Ln[x(x-1)T+1] / (1-x) then apply l'hopitals,

Rewrite Ln[x(x-1)T+1]=a (1-x)=b

Lim X->1 =da/dx / db/dx

da/dx = T(2x-1)/[Tx(x-1)+1]

db/dx = -1

da/dx / db/dx = -T(2x-1)/[Tx(x-1)+1] = -T

ln(y) = -T

Y= e^-T

Sorry for many edits !

Yup, i edit my posts constantly too so i do the same thing to everyone