limit of indeterminate powers w/ natural log

[wranglr88]

hey guys, just got stuck on a little limit here, and hoping you guys could give me a hand real quick.
here's the problem:

lim (x/x+1)^x
x->inf

i figured:
y = (x/x+1)^x
lny=xln(x/x+1)
lim lny = lim (ln{x/x+1}/x^-1)
x->inf x->inf
.....
so i was thinking of using l'hopital's rule on this, but got stuck on how to find the derivative of the natural log part.

thanks for the help!

 

[BigGlenntheHeavy]

\(\displaystyle Let \ y \ = \ \lim_{x\to\infty}\bigg(\frac{x}{x+1}\bigg)^x\)

\(\displaystyle Then, \ ln|y| \ = \ ln\bigg[\lim_{x\to\infty}\bigg(\frac{x}{x+1}\bigg)^x\bigg], \ taking \ the \ natural \ logarithm \ of \ both \ sides\)

\(\displaystyle Hence, \ ln|y| \ = \ \lim_{x\to\infty}\bigg[ln\bigg(\frac{x}{x+1}\bigg)^x\bigg], \ function \ is \ continuous\)

\(\displaystyle = \ \lim_{x\to\infty}\bigg[xln\bigg(\frac{x}{x+1}\bigg)\bigg] \ = \ \lim_{x\to\infty}\frac{ln\bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}\)

\(\displaystyle Note: \ The \ above \ gives \ the \ indeterminant \ form \ \frac{0}{0}\)

\(\displaystyle Therefore, \ ln|y| \ = \ \lim_{x\to\infty}\frac{-x}{x+1} \ = \ \lim_{x\to\infty}\frac{-1}{1} \ = \ -1\)

\(\displaystyle Now. \ ln|y| \ = \ -1 \ \implies \ y \ = \ e^{-1}, \ QED\)

\(\displaystyle Also \ note \ that \ the \ Marqui \ was \ used \ three \ times.\)

 

[galactus]

Oh well, I will give my two cents for what it's worth.

Rewrite:
\(\displaystyle y=\lim_{x\to {\infty}}\left(1-\frac{1}{x+1}\right)\)

Log on both sides and rewrite using log laws:
\(\displaystyle ln(y)=\lim_{x\to 0}\left[\frac{ln(1-\frac{1}{x+1})}{\frac{1}{x}}\right]\)

Now, let \(\displaystyle t=\frac{1}{x}, \;\ \frac{1}{x+1}=\frac{t}{t+1}\)

Noting that as \(\displaystyle x\to {\infty}\), then \(\displaystyle t\to 0\)

\(\displaystyle \lim_{t\to 0}\frac{ln(\frac{1}{t+1})}{t}\)

\(\displaystyle =\lim_{t\to 0}\frac{-ln(t+1)}{t}\)

Let \(\displaystyle h=ln(1+t)\)

Then, we have \(\displaystyle \lim_{h\to 0}\frac{-h}{e^{h}-1}\)

Make the observation that this is the same as \(\displaystyle \frac{-1}{\frac{e^{h}-1}{h}}\)

The denominator is the definition of the derivative of e at x=0. This is 1

Since we have \(\displaystyle \frac{-1}{e^{0}}\), we have the limit of -1.

But, from the beginning we used ln of both sides, so we get \(\displaystyle e^{-1}=\frac{1}{e}\) as the limit.

 

[Deleted member 4993]

\(\displaystyle Let \ y \ = \ \lim_{x\to\infty}\bigg(\frac{x}{x+1}\bigg)^x\)

\(\displaystyle \ \ \\ = \ \lim_{x+1 \to\infty}\left [ \bigg(1 - \frac{1}{x+1}\bigg)^{x+1}\cdot \frac{1}{1 - \frac{1}{x+1}}\right ]\)

\(\displaystyle = e^{-1} \cdot 1\)