Asymptote to the curve $y= \dfrac {3x}{2}\log\left(\mathrm e-\frac{1}{3x}\right)$ $$y= \dfrac {3x}{2}\log\left(\mathrm e-\frac{1}{3x}\right)$$
We have to find the asymptote to the curve I tried solving the problem by taking its limit as $x$ tends to infinity. But I failed. Can anyone help.
 A: As $x \to \infty$, we see that $\frac{1}{3x} \to 0$ and so $\log\left(\mathrm e -\frac{1}{3x}\right) \to 1$.
As $x \to \infty$, $\frac{3x}{2} \to \infty$.
Hence, $\frac{3x}{2}$ dominates $\log\left(\mathrm e -\frac{1}{3x}\right)$ as $x \to \infty$.
It follows that $\frac{3x}{2}\log\left(\mathrm e -\frac{1}{3x}\right)$ behaves like $\frac{3x}{2} \times 1$ when $x \to \infty$, i.e. the asymptote is $y=\frac{3x}{2}$.
A: The asymptote is in the form $y=m\,x+b$. First compute $m$:
$$
m=\lim_{x\to\infty}\frac{\dfrac{3\,x}{2}\,\log\Bigl(e-\frac{1}{3\,x}\Bigr)}{x}=\frac32.
$$
As for $m$:
$$
m=\lim_{x\to\infty}\Bigl(\frac{3\,x}{2}\,\log\Bigl(e-\frac{1}{3\,x}\Bigr)-\frac{3\,x}{2}\Bigr)=\lim_{x\to\infty}\frac{3\,x}{2}\,\log\Bigl(1-\frac{1}{3\,e\,x}\Bigr)=-\frac{1}{2\,e}.
$$
The asymptote is
$$
y=\frac{3\,x}{2}-\frac{1}{2\,e}.
$$
A: The slope of the asymptote depends on the actual base of our "log". The following figure depicts the function for base $e$ and for base $10$.

Indeed, let the actual base be $a$ then 
$$Y= \dfrac {3x}{2}\log_a\left(\mathrm e-\frac{1}{3x}\right)=\frac{3x}{2\ln(a)}\ln\left(\mathrm e-\frac{1}{3x}\right).$$
Now, one can repeat the argumentation of the other answers and the slope of the asymptote turns out to be:
$$\frac{3}{2\ln(a)}.$$
A: \begin{align}
&{3x \over 2}\ln\left(\mathrm{e} - {1 \over 3x}\right) =
{3 \over 2}\,x\left[1 + \ln\left(1 - {1 \over 3\mathrm{e}x}\right)\right] =
{3 \over 2}\,x\left[1 - {1 \over 3\mathrm{e}x} + \mathrm{O}\left(1 \over x^{2}\right)\right]
\\[5mm] = &\
\color{#f00}{{3 \over 2}\,x  - {1 \over 2\mathrm{e}}} +
\mathrm{O}\left(1 \over x\right)\quad\mbox{as}\quad x \to \infty
\end{align}
