# L'Hopital's limit

• Apr 27th 2011, 06:18 PM
mrantoine
L'Hopital's limit
$\lim_{x \to \infty}$ (4^x + 7^x)^(1/x)

First I took the ln of (4^x + 7^x)^(1/x) to get (4^x + 7^x)/x and derived the top and bottom, so I got a function of
(4^x *ln4 + 7^x *ln7)/(4^x + 7^x)

and I took the derivative of the top and bottom again, but I could tell already that things were going nowhere :/
• Apr 27th 2011, 06:41 PM
Ackbeet
Hmm. Let's see.

$\ln\left(\lim_{x\to\infty}(4^{x}+7^{x})^{1/x}\right)=\lim_{x\to\infty}\ln\left((4^{x}+7^{x})^ {1/x}\right)=\lim_{x\to\infty}\left[\frac{\ln(4^{x}+7^{x})}{x}\right]$
$=\underbrace{\lim_{x\to\infty}\left[\frac{4^{x}\ln(4)+7^{x}\ln(7)}{4^{x}+7^{x}}\:\frac {1}{1}\right]}_{\text{L'Hopital's Rule}}.$

There is where you ended. At this point I would divide top and bottom by $7^{x},$ instead of using L'Hopital's Rule again.
• Apr 27th 2011, 06:51 PM
mrantoine
ohhhhhhhh i see now. Dividing the top and bottom by 7^x yields $\lim_{x \to \infty}$ ((((4^x)*ln4) /7^x) + ln7)/((4^x/7^x)+1) and 4^x*ln4/7^x and 4^x / 7^x both go to zero since 7^x will grow at a faster rate than 4^x, so ln $\lim_{x \to \infty}$ (4^x + 7^x)^(1/x) = e^(ln7) = 7

Thank you so much!

But real quick: Why did you have to divide by 7^x to get the right answer?
• Apr 27th 2011, 06:52 PM
Ackbeet
You got it, and you're welcome!