Limit at infinity.

Question

$$\lim _{p\to \infty }\left(p\ln \left(e\left(1+\frac{1}{p}\right)^{1-p}\right)\right)?$$

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?

Answer 1

Hint: The expression equals

$$p + p(1-p)\ln(1+1/p).$$

Recall $\ln (1+u) = u-u^2/2 + O(u^3)$ as $u\to 0.$

Answer 2

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}