Limit using L'hospital and maybe another trick

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May 25th 2010, 02:10 PM
odbg9005

Limit using L'hospital and maybe another trick

Hi, I've been sitting on this more than an hour and still can't find the solution :(

$\displaystyle\lim_{x \to{\infty}}{\displaystyle{x(}(1+\frac{1}{x})^{x} - {e})}$

I put this in a function grapher and it looks like the result should be ${-}\frac{e}{2}$ , but I don't understand why this is true or how do I get to this solution. I assume that L'hospital should be used and supposedly there's another trick?

Thanks in advance
May 25th 2010, 02:50 PM
AllanCuz

Quote:

Originally Posted by odbg9005

Hi, I've been sitting on this more than an hour and still can't find the solution :(

$\displaystyle\lim_{x \to{\infty}}{\displaystyle{x(}(1+\frac{1}{x})^{x} - {e})}$

I put this in a function grapher and it looks like the result should be ${-}\frac{e}{2}$ , but I don't understand why this is true or how do I get to this solution. I assume that L'hospital should be used and supposedly there's another trick?

Thanks in advance

Yikes!
May 25th 2010, 02:57 PM
galactus

We can use L'Hopital.

Once, and get:

$-\lim_{x\to {\infty}}\frac{x^{2}\left(xln(\frac{x+1}{x})+ln(\f rac{x+1}{x})-1\right)\left(\frac{x+1}{x}\right)^{x}}{x+1}$

L'Hopital again:

$-\frac{1}{2}\lim_{x\to {\infty}}\left(\frac{x+1}{x}\right)^{x}\cdot\frac{ x^{2}}{(1+x)^{2}}$

Rewrite:

$\frac{-1}{2}\lim_{x\to {\infty}}\frac{1}{1+\frac{2}{x}+\frac{1}{x^{2}}}\c dot\lim_{x\to {\infty}}\left(\frac{1+x}{x}\right)^{x}$

Now, we can see that the right side is the famous e limit, the center one is 1 and we get $\frac{-e}{2}$
May 25th 2010, 03:39 PM
odbg9005

Thank you so much for the reply. However, I still don't understand how you did the second activation of L'Hospital. I tried to calculate the derivative of the numerator, and it seems insane, like it will take months to complete and require a whole notebook. Is there something I'm missing here?
May 26th 2010, 02:30 AM
galactus

I must admit, I simplified this with tech...not by hand. But, if you use the product rule for limits you can work it out.

That is, the limit of the product is the product of the limit.

$-\lim_{x\to {\infty}}\frac{x^{2}\left(xln(\frac{x+1}{x})+ln(\f rac{x+1}{x})-1\right)\lim_{x\to {\infty}}\left(\frac{x+1}{x}\right)^{x}}{x+1}$

Along with the log laws, it should simplify down.

Remember that $\lim_{x\to {\infty}}xln(\frac{x+1}{x})=1$ , because it is ln(e)=1

Like I said, I did not work through this by hand. Too lazy. (Wink)
May 26th 2010, 06:51 AM
nahduma

Simplifying the derivation

Actually, you can greatly simplify the derivation if you put y=1/x in the limit.
so as $x\to \inf , \\ y\to 0$
the equation becomes
$<br /> \lim_{y\to0} \frac{(1+y)^{1/y}-e}{y}$
applying l'hospital's rule, we get
$\lim_{y\to0} (1+y)^{1/y} \frac{y-(1+y)ln(1+y)}{y^2}$ which is
$\lim_{y\to0} (1+y)^{1/y} \lim_{y\to0} \frac{y-(1+y)ln(1+y)}{y^2}$
Since the left hand side is e, we apply l'hospital again to the right side
$e \lim_{y\to0} \frac{-ln(1+y))}{2y}$
Applying l'hospital again, we get
$\frac{-e}{2}$