L'Hospital's Rule and indeterminate powers What is $\displaystyle \lim_{x\to\infty}\left(\frac{17x}{17x+9}\right)^{3x}$?
I tried to solve this problem and could not understand this. 
I know that it is an exponential equation of the type $\infty^{\infty}$. I got to $3x (\ln 17x - \ln (17x+9) = \ln y$. 
How do I proceed from there? Am I doing this incorrectly in the first place? 
 A: Note that
$$
\begin{align*}
\lim_{x \to \infty} \left(\frac{17x}{17x+9}\right)^{3x} &= \lim_{x\to\infty}\left(\dfrac{1}{1+\frac{9}{17x}}\right)^{3x} = \lim_{x\to\infty}\left[\dfrac{1}{\left(1+\frac{9}{17x}\right)^{17x/9}}\right]^{27/17}\\
&= \left[\dfrac{1}{\lim\limits_{x\to\infty} \left(1 + \frac{9}{17 x}\right)^{17x/9}}\right]^{27/17} = \left(\dfrac{1}{e}\right)^{27/17} = \dfrac{1}{e^{27/17}}
\end{align*}
$$
A: Slow down: If I am reading/interpreting your function correctly, we  have $$\lim_{x\to \infty} \left(\dfrac{17x}{17x+9}\right)^{3x}$$
is not of the form $\infty^\infty$.  We have 
$$ \left(\dfrac{17x}{17x+9}\right)^{3x} \to 1^\infty \;\text{ as}\;\; x\to \infty$$
Since
$$
\begin{align}
\lim_{x\to \infty} \left(\dfrac{17x}{17x+9}\right)^{3x} 
& = \lim_{x\to \infty} \left(1 - \dfrac{9}{17x + 9}\right)^{3x}
\end{align}
$$
A: Basically
$$\lim_{x\to\infty} \left(\frac{17x}{17x+9}\right)^{3x} = \lim_{x\to\infty}\left(1-\frac{9}{17x+9}\right)^{3x} $$ $$= \lim_{x\to\infty}\left(1-3\frac{9}{17x+9}+3\left(\frac{9}{17x+9}\right)^2+\left(\frac{9}{17x+9}\right)^3\right)^{x} $$ $$= \lim_{x\to\infty}\left(1-3\frac{9}{17x+9}\right)^{x} = \lim_{x\to\infty}\exp\left(x\log\left(1-\frac{27}{17x+9}\right)\right) $$ $$= \lim_{x\to\infty} \exp\left(x\left(-\frac{27}{17x+9}\right)\right) = \lim_{x\to\infty} \exp\left(-\frac{27x}{17x+9}\right) = e^{-\frac{27}{17}}.$$
A: To carry it out the way you were approaching it, taking the natural logarithm of the expression and applying the appropriate "limit law" leads to
$$\ln \lim_{x \rightarrow \infty} y  =  \lim_{x \rightarrow \infty} \ln y=  \lim_{x \rightarrow \infty} 3x \cdot [ \ln (17x) - \ln (17x + 9) ] ,$$ 
which is now an "indeterminate product" limit with an included "indeterminate difference" factor.  (This is what you had so far.)  Generally, to make this ready for l'Hopital by forming an "indeterminate ratio", it is best to place logarithmic terms in the numerator.  From here,
$$\ln \lim_{x \rightarrow \infty} y  =  \lim_{x \rightarrow \infty} \frac{[ \ln (17x) - \ln (17x + 9) ]}{\frac{1}{3x}} ,$$
for which "direct substitution" gives us the result $\frac{0}{0}$ (it is not hard to show that the limit of the difference is zero). We are now ready to apply the LHR:
$$\ln \lim_{x \rightarrow \infty} y  =  \lim_{x \rightarrow \infty} \frac{[ \ln (17x) - \ln (17x + 9) ] '}{(\frac{1}{3x})'} = \lim_{x \rightarrow \infty} \frac{ (\frac{1}{x}) - \frac{17}{(17x + 9)}}{(-\frac{1}{3x^2})}$$ 
[$\frac{d}{dx} \ln(kx) = \frac{d}{dx} [\ln(k) + \ln(x)]  =  0 + \frac{1}{x}$]
$$= \lim_{x \rightarrow \infty} -\frac{(3x^2)[(17x + 9) - 17x]} {x(17x + 9)} = \lim_{x \rightarrow \infty} -\frac{(3x^2) \cdot 9} {17x^2 + 9x} = -\frac{27} {17} .$$
So, at last, $\ln \lim_{x \rightarrow \infty} y  =  -\frac{27} {17} \Rightarrow \lim_{x \rightarrow \infty} y = e^{-27/17} $.  Your approach works, but requires some care and patience (I had to catch my "arithmetic mistakes" a couple times...)
As this and the demonstrations of the other responders suggest, the limit at infinity of a power of a rational function will be some power of $e$ .
