TITLE: Limit at infinity.
QUESTION [5 upvotes]: $$\lim _{p\to \infty }\left(p\ln \left(e\left(1+\frac{1}{p}\right)^{1-p}\right)\right)?$$
I was getting $1$ as an answer but answer is $1.5$. 
I solved it like this:-
$$\lim_{p\to \infty} p\ln\left(e\left(\left(1+\frac{1}{p}\right)^{\left(1-p\right)\frac{p}{p}}\right)\right)$$
$$\lim_{p\to \infty} p\ln\left(e\cdot e^{\frac{1-p}{p}}\right)$$
$$\lim_{p\to \infty} p\ln\left(e^{\frac{1}{p}}\right)$$
$$\lim_{p\to \infty}p\frac{1}{p} = 1$$
What am I doing wrong?

REPLY [3 votes]: First observe that
\begin{align}
p \ln \left[ e \left(1+\frac1p\right)^{1-p} \right]
    & = p \left[ 1 + (1-p)\ln \left(1+\frac1p\right) \right] \\
    & = p + (p-p^2) \ln\left(1+\frac1p\right)
\end{align}
Now, recall the Taylor series expansion of $\ln (1+x)$ around $x = 0$:
$$
\ln (1+x) = x - \frac{x^2}{2} + o(x^2)
$$
So then we have
\begin{align}
\lim_{p \to \infty} p \ln \left[ e \left(1+\frac1p\right)^{1-p} \right]
    & = \lim_{p \to \infty} p + (p-p^2) \ln\left(1+\frac1p\right) \\
    & = \lim_{p \to \infty} p + (p-p^2) \left[\frac1p-\frac{1}{2p^2} + o\left(\frac{1}{p^2}\right)\right] \\
    & = \lim_{p \to \infty} p + 1 - p + \frac12 + o(1) \\
    & = \frac32
\end{align}