Limit using L'hospital and maybe another trick

May 2010
2
0
Hi, I've been sitting on this more than an hour and still can't find the solution :(

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

I put this in a function grapher and it looks like the result should be \(\displaystyle {-}\frac{e}{2}\), 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
 
Apr 2010
384
153
Canada
Hi, I've been sitting on this more than an hour and still can't find the solution :(

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

I put this in a function grapher and it looks like the result should be \(\displaystyle {-}\frac{e}{2}\), 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!
 

galactus

MHF Hall of Honor
Jul 2006
3,002
1,124
Chaneysville, PA
We can use L'Hopital.

Once, and get:

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

L'Hopital again:

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

Rewrite:

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

Now, we can see that the right side is the famous e limit, the center one is 1 and we get \(\displaystyle \frac{-e}{2}\)
 
May 2010
2
0
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?
 

galactus

MHF Hall of Honor
Jul 2006
3,002
1,124
Chaneysville, PA
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.

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

Along with the log laws, it should simplify down.

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

Like I said, I did not work through this by hand. Too lazy. (Wink)
 
Feb 2010
26
1
Simplifying the derivation

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