Find all asymptotes of y=ln (e+ dfrac1x)^x=x ln (e+

Find all asymptotes of $y=\ln\!\left(e+\dfrac1x\right)^{\!x}=x\ln\!\left(e+\dfrac1x\right)$.

Step 1: No asymptote at $x=0$

As $x\to0^+$, $x\ln(e+1/x)\sim x\ln(1/x)=-x\ln x\to0$: no vertical asymptote there.

Step 2: Vertical asymptote where the argument vanishes

The argument $e+1/x$ vanishes at $x=-1/e$; approaching from the right, \[\lim_{x\to(-1/e)^{+}} x\ln\!\left(e+\frac1x\right) = -\frac1e\cdot(-\infty) = +\infty \;\Longrightarrow\; x=-\frac1e. \tag{1}\]

Step 3: Oblique asymptote — slope

\[ k = \lim_{x\to\pm\infty}\ln\!\left(e+\frac1x\right) = \ln e = 1. \tag{2} \]

Step 4: Oblique asymptote — intercept

Since $\ln(1+u)\sim u$ as $u\to0$: \[b = \lim_{x\to\pm\infty} x\ln\!\left(1+\frac1{ex}\right) \sim \lim_{x\to\pm\infty} x\cdot\frac1{ex} = \frac1e. \tag{3}\]

Step 5: Verification

\[x=1000\text{: } y-(x+1/e)\approx-1.8\times10^{-7}\to0\ \checkmark\]