L'Hospital's Rule

**skabani** · September 9th 2008, 02:23 PM

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!!

**11rdc11** · September 9th 2008, 02:29 PM

I think the second one is the number e but just looked at it real fast

**Moo** · September 10th 2008, 06:39 AM

Hello,

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

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

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...

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