# Thread: L'Hopital's Rule and Improper Integrals

1. ## L'Hopital's Rule and Improper Integrals

$\lim (e^x+x)^\frac{2}{x}$
$x->0^+$

Describe the type of indeterminate form that is obtained by direct substitutions.
Evaluate the integral using L'Hopital's Rule if necessary.

I understand that indeterminate means having no definite or definable value. But I don't what to put for it.

2. lt(x-->0+)(e^x+x)^2/x

p=(e^x+x)^2/x

so,ln p=2ln(e^x+x)/x

so,lim ln p=2 lim (x-->0+)ln (e^x+x)/x

so,ln lim p=2 lim(x-->0+)1/(e^x+x)[by l'hospital's rule]

so,lim p=e^2

3. Originally Posted by saiyanmx89
$\lim (e^x+x)^\frac{2}{x}$
$x->0^+$

Describe the type of indeterminate form that is obtained by direct substitutions.
Evaluate the integral using L'Hopital's Rule if necessary.

I understand that indeterminate means having no definite or definable value. But I don't what to put for it.
Actually when dealing with limits, the indeterminate forms are

$\frac{0}{0}$ or $\frac{\infty}{\infty}$

So which form do you get at the moment?

4. The limit equals $e^4.$ Here's my approach:

$\left( e^{x}+x \right)^{\frac{2}{x}}=\left\{ e^{x}\bigg( 1+\frac{x}{e^{x}} \bigg) \right\}^{\frac{2}{x}}=e^{2}\left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}},$ thus we only need to prove that $\left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}\to e^{2}$ as $x\to0.$

We have $\left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}=\exp \left\{ \frac{2}{e^{x}}\cdot \ln \bigg( 1+\frac{x}{e^{x}} \bigg)\cdot \frac{e^{x}}{x} \right\},$ and $\ln \left( 1+\frac{x}{e^{x}} \right)\cdot \frac{e^{x}}{x}\to 1$ as $x\to0.$

(Just we need to recall that $\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1.$)

Finally $\left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}\to e^{2}$ as $x\to0,$ as required and the whole limit tends to $e^4.\quad\blacksquare$

the error here is with the chain rule...
you were missing the e^x+1--->2 as x-->0
which is why you only got e^2 and not e^4.

Originally Posted by sbcd90
lt(x-->0+)(e^x+x)^2/x

p=(e^x+x)^2/x

so,ln p=2ln(e^x+x)/x

so,lim ln p=2 lim (x-->0+)ln (e^x+x)/x

so,ln lim p=2 lim(x-->0+)(e^x+1)/(e^x+x)[by l'hospital's rule]

so,lim p=e^[(2)(2)]=e^4

6. Originally Posted by Prove It
Actually when dealing with limits, the indeterminate forms are

$\frac{0}{0}$ or $\frac{\infty}{\infty}$

So which form do you get at the moment?
$0^0$ ?

7. Originally Posted by saiyanmx89
$0^0$ ?

whener you have $0^0$
you just call that $y=0^0$
and then proceed via $\ln y$...

8. Originally Posted by Prove It
Actually when dealing with limits, the indeterminate forms are

$\frac{0}{0}$ or $\frac{\infty}{\infty}$

So which form do you get at the moment?
$0^0$ and $\infty^\infty$ are also "indeterminate forms" and, I think, more relevant here.