# L'Hospital's Rule

• September 9th 2008, 03:23 PM
skabani
L'Hospital's Rule
use l'hospital's rule to find the limit:

1. lim (tan 2x)^x
x-> 0+

i got 1 using ln y, but im not sure if its right..

2. lim (1+ (a/x))^bx
x-> INFINITY

again i got 1...but am unsure

any help is appreciated...thanks!!
• September 9th 2008, 03:29 PM
11rdc11
I think the second one is the number e but just looked at it real fast
• September 10th 2008, 07:39 AM
Moo
Hello,
Quote:

Originally Posted by skabani
use l'hospital's rule to find the limit:

1. lim (tan 2x)^x
x-> 0+

i got 1 using ln y, but im not sure if its right..

I get 1 too :)

Quote:

2. lim (1+ (a/x))^bx
x-> INFINITY
Quote:

Originally Posted by 11rdc11
I think the second one is the number e but just looked at it real fast

Substitute $t=\frac xa$
Thus $bx=abt$ and $\frac ax=\frac 1t$

$\lim_{x \to \infty} \left(1+\frac ax\right)^{bx}=\lim_{t \to \infty} \left(1+\frac 1t\right)^{t \cdot (ab)}$

Remember the rule $(a^b)^c=(a^c)^b=a^{bc}$

Therefore :

$\lim_{t \to \infty} \left(1+\frac 1t\right)^{t \cdot (ab)}=\lim_{t \to +\infty} \left(\left(1+\frac 1t\right)^t\right)^{ab}=e^{ab} \quad (*)$

$(*) \lim_{t \to + \infty} \left(1+\frac 1t\right)^t=e$ (by taking the logarithm and applying l'Hospital's rule)
(a little substitution can be done if t tends to $-\infty$)

Otherwise, applying the logarithm :

$\ln(y)=\lim_{x \to \infty} bx \left(1+\frac ax\right)$

Then apply l'Hospital's rule...