# Thread: Limit using L'Hospitals Rule

1. ## Limit using L'Hospitals Rule

lim (1+C/x)^(hx)
x-> infinity

I have to evaluate a limit

since if you plug in infinity/infinity you have to use L'Hospitals rule

Thank you

2. Originally Posted by Zvaigzdute
lim (1+C/x)^(hx)
x-> infinity

I have to evaluate a limit

since if you plug in infinity/infinity you have to use L'Hospitals rule

Thank you
$\displaystyle \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{c}{x}} \right)^{hx}} = \mathop {\lim }\limits_{x \to \infty } \exp \ln {\left( {1 + \frac{c}{x}} \right)^{hx}} = \mathop {\lim }\limits_{x \to \infty } \exp \left\{ {hx\ln \left( {1 + \frac{c}{x}} \right)} \right\} =$

$\displaystyle = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\ln \left( {1 + \frac{c}{x}} \right)}}{{\frac{1}{{hx}}}} = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{d}{{dx}}\ln \left( {1 + \frac{c}{x}} \right)}}{{\frac{d}{{dx}}\frac{1}{{hx}}}} = \mathop {\lim }\limits_{x \to \infty } \exp \frac{ch}{{1 + \frac{c}{x}}} =e^{ch}.$

3. Hello, Zvaigzdute!

Another approach . . .

Evaluate: .$\displaystyle \lim_{x\to\infty}\left(1+\frac{C}{x}\right)^{hx}$
Let: $\displaystyle y \;=\;\left(1 + \frac{C}{x}\right)^{hx}$

Take logs: .$\displaystyle \ln y \;=\;\ln\left(1 + \frac{C}{x}\right)^{hx} \;=\;hx\cdot\ln\left(1 + \frac{C}{x}\right) \;=$ .$\displaystyle \frac{h\ln\left(1 + \frac{C}{x}\right)}{\frac{1}{x}} \;=\;\frac{h\ln\left(1 + Cx^{-1}\right)}{x^{-1}}$

Apply l'Hopital: .$\displaystyle \frac{h\cdot\dfrac{1}{1+Cx^{-1}}\cdot(-Cx^{-2})} {-x^{-2}} \;=\;\frac{Ch}{1 + \frac{C}{x}}$

Then: .$\displaystyle \lim_{x\to\infty}\ln y \;=\;\lim_{x\to\infty} \left(\frac{Ch}{1 + \frac{C}{x}}\right) \;=\;\frac{Ch}{1+0} \;=\;Ch$

Therefore: .$\displaystyle \ln y \:=\:Ch \quad\Rightarrow\quad y \;=\;e^{Ch}$