# Math Help - L'Hopital's Rule question...

1. ## L'Hopital's Rule question...

Hi, I have two questions requiring L'Hopital's Rule that I can't seem to get. They are

And

For the first one I didn't even think the rule was necessary. I thought I could just plug in infinity and get (1+0)^INF. But this was obviously wrong. For the second I thought I should try to get it into a quotient problem, but I kept getting 0 in the denominator. I'd really appreciate help. Thanks!

2. let $y = \ln \Big(1 + \frac{9}{x}\Big) ^{\frac{x}{2}} = \frac{x}{2} \ln \Big(1 + \frac{9}{x}\Big)$ $= \frac{\ln \Big(1 + \frac{9}{x}\Big)}{\frac{2}{x}}$

which is the indeterminate form $\frac{0}{0}$

so $\lim_{x \to \infty} y = \lim_{x \to \infty} \frac{\ln \Big(1 + \frac{9}{x}\Big)}{\frac{2}{x}}$

$= \lim_{x \to \infty} \frac{\frac{1}{1+\frac{1}{9x}}* -\frac{9}{x^{2}}}{-\frac{2}{x^{2}}} = \frac{9}{2} \lim_{x \to \infty} \frac{1}{1+\frac{1}{9x}} = \frac{9}{2}*1 = \frac{9}{2}$

the original function is $e^{y}$

so $\lim_{x \to \infty} \Big(1 + \frac{9}{x}\Big) ^{\frac{x}{2}} = \lim_{x \to \infty} e^{y} = e^{\frac{9}{2}}$