Compute $\lim_{x\rightarrow \infty}(\frac{ax+b}{ax+c})^{rx}$ Compute $\lim_{x\rightarrow \infty} (\frac{ax+b}{ax+c})^{rx}$
I try taking the natural log, which is $\ln(y)=rx\ln(\lim_{x\rightarrow \infty} \frac{ax+b}{ax+c})$ but that turns into  $0x\infty$ no matter how many times you do l'hospitals
The answer, from wolfram alpha, is e^(b-c)r/a. 
I also tried using L'H for $(\ln(\frac{ax+b}{ax+c})/rx^{-1})$. This resulted in something like $-r(c-b).$
Any help would be great, thanks.
 A: You are on the right track with this:
$$ \text{ln}(y)=\lim_{x\to\infty}rx\text{ln}(\frac{ax+b}{ax+c}). $$
Let us transform it a little more:
$$
\lim_{x\to\infty}rx\text{ln}(\frac{ax+b}{ax+c})=\lim_{x\to\infty}rx\text{ln}(1+\frac{b-c}{ax+c}) = \text{Using the Taylor formula for ln(1+x)}=\\=\lim_{x\to\infty}rx*(\frac{b-c}{ax+c} )+ O(\frac{1}{x})=(b-c)\frac{r}{a}.
$$
Exponentiating the answer will then give you the desired limit:
$$
y=e^{(b-c)\frac{r}{a}}.
$$
A: $\lim_{x\rightarrow \infty}(\frac{ax+b}{ax+c})^{rx}= \lim_{x\rightarrow \infty}(\frac{ax+c+(b-c)}{ax+c})^{rx} =\lim_{x\rightarrow \infty}(1+\frac{b-c}{ax+c})^{rx} =\lim_{x\rightarrow \infty}(1+\frac{b-c}{ax+c})^{\frac{ax+c}{b-c}\frac{(b-c)rx}{ax+c}}= \lim_{x\rightarrow \infty}(1+\frac{b-c}{ax+c})^{\frac{ax+c}{b-c}\lim_{x\rightarrow \infty}\frac{(b-c)rx}{ax+c}}=e^{(b-c)r/a}$
For the second limit: $\lim_{x\rightarrow \infty}\frac{(b-c)rx}{ax+c}= \lim_{x\rightarrow \infty}\frac{(b-c)r}{a+\frac{c}{x}}=$ (if $x\rightarrow \infty\Rightarrow \frac{c}{x}\rightarrow 0)={(b-c)r/a}$.
A: We can write the limit as $\left( 1 + \frac{b-c}{ax+c} \right)^{rx}$ which is the same as $ \left( \left( 1 + \frac{b-c}{ax+c} \right)^{ax+c} \right)^\frac{rx}{ax+c}$. The inner limit is $e^{b-c}$ and the exponent tends to $\frac{r}{a}$. Putting them together gives you the answer that Wolfram Alpha returns.
