# limits(l'hopital's rule)

• Apr 16th 2008, 02:59 AM
ashes
limits(l'hopital's rule)
$\displaystyle lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x$ therefore i applied l'hopital s rule to make $\displaystyle lim_{x\rightarrow{\infty}}\frac{-(x+1)^{x-1}}{x^2}$ I hope my differentiation is correct. So do i keep differentiating till I remove the x term from the denominator before i apply the limit?
• Apr 16th 2008, 03:24 AM
mr fantastic
Quote:

Originally Posted by ashes
$\displaystyle lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x$ therefore i applied l'hopital s rule to make $\displaystyle lim_{x\rightarrow{\infty}}\frac{-(x+1)^{x-1}}{x^2}$ I hope my differentiation is correct. So do i keep differentiating till I remove the x term from the denominator before i apply the limit?

I'm sorry but you're differentiations are terribly wrong. If you show how you did them I'll point out the many salto mortales .....

You actually have a standard form:

$\displaystyle \lim_{x\rightarrow{\infty}}\left(\frac{x+1}{x}\rig ht)^x = \lim_{x\rightarrow{\infty}}\left(1 + \frac{1}{x}\right)^x = e$

by definition.
• Apr 16th 2008, 05:04 AM
colby2152
Quote:

Originally Posted by ashes
$\displaystyle lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x$ therefore i applied l'hopital s rule to make $\displaystyle lim_{x\rightarrow{\infty}}\frac{-(x+1)^{x-1}}{x^2}$ I hope my differentiation is correct. So do i keep differentiating till I remove the x term from the denominator before i apply the limit?

Mr. Fantastic's answer is correct, but I have one point to make...

All too common mistake... you cannot derive $\displaystyle x^x$ via basic powers rule and get $\displaystyle x*x^{x-1}$. Notice that the result simplifies back to $\displaystyle x^x$

You need to use the natural log and implicit differentiation to solve for such a derivative.

$\displaystyle y=x^x$

Take natural log of both sides...

$\displaystyle \ln|y|=x \ln|x|$

$\displaystyle \frac{1}{y}y' = \ln|x| + 1$

Derive with respect to x on both sides

$\displaystyle y' = y(\ln|x| + 1)$

Substitute for y from original equation

$\displaystyle y' = x^x(\ln|x| + 1)$
• Apr 16th 2008, 05:42 AM
Isomorphism
Interestingly we were taught to do this with a different perspective.
$\displaystyle d(y^x) = x.y^{(x-1)} dy + y^x \ln |y| dx$

Now substitute y=x

$\displaystyle d(x^x) = x.x^{(x-1)} dx + x^x \ln |x| dx$

$\displaystyle d(x^x) = x^{x} dx + x^x \ln |x| dx$

$\displaystyle \frac{d(x^x)}{dx} = x^{x} + x^x \ln |x|$