I cannot solve this limit $$
\lim_{n\to\infty}\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}=e^{b-\frac{1}{2}}
$$
I am doing it like this, and I cannot find the mistake:
$$
\lim_{n\to\infty}\frac{1}{e^n}e^{n+b+c/n}=
\lim_{n\to\infty}e^{n+b-n+c/n}=e^b
$$
 A: I would say that $$\log_e\left(\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}\right) $$  $$ = (bn+c+n^2)\log_e\left(\frac{1}{n}+1\right) - n$$  $$= (bn+c+n^2)\left(\frac{1}{n}-\frac{1}{2\,{n}^{2}}+\frac{1}{3\,{n}^{3}}-\frac{1}{4\,{n}^{4}}+\cdots\right) - n$$ $$= b - \frac{b}{2n} + o(1/n) +\frac{c}{n} - o(1/n) +n -\frac{1}{2} + \frac{1}{3n} -o(1/n) -n$$ $$= b  -\frac{1}{2} +\frac{-3b+6c+2}{6n} + o(1/n)$$
making the limit of the logarithm $b-\frac{1}{2}$ as $n\to \infty$ and so the original limit $\displaystyle e^{b-\frac{1}{2} }$.
A: You could use L'Hospital's rule and $x=\frac{1}{n}$ to get
$\displaystyle\lim_{n\to\infty}\ln\bigg[\frac{(1+\frac{1}{n})^{bn+n^2+c}}{e^n}\bigg]=\lim_{n\to\infty}(bn+n^2+c)\ln\big(1+\frac{1}{n}\big)-n$
$=\displaystyle\lim_{x\to0}\bigg(\frac{b}{x}+c+\frac{1}{x^2}\bigg)\ln(1+x)-\frac{1}{x}=\lim_{x\to0}\frac{(bx+cx^2+1)\ln(1+x)-x}{x^2}$
$=\displaystyle\lim_{x\to0}\frac{\ln(1+x)-\frac{x}{bx+cx^2+1}}{\frac{x^2}{bx+cx^2+1}}=\displaystyle\lim_{x\to0}\frac{\frac{1}{1+x}-\frac{1-cx^2}{(bx+cx^2+1)^2}}{\frac{bx^2+2x}{(bx+cx^2+1)^2}}=\lim_{x\to0}\frac{(bx+cx^2+1)^2-(1+x)(1-cx^2)}{(1+x)(bx^2+2x)}$
$\displaystyle=\lim_{x\to0}\frac{b^2x+2b+2bcx^2+3cx+c^2x^3+cx^2-1}{(1+x)(bx+2)}=\frac{2b-1}{2}=b-\frac{1}{2}$,
so $\displaystyle\lim_{n\to\infty}\bigg(\frac{(1+\frac{1}{n})^{bn+n^2+c}}{e^n}\bigg)=e^{b-\frac{1}{2}}$.
A: 
I am doing it like this, and I cannot find the mistake

Note that $\left(1+\frac 1n\right)^n\not=e$ and that
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n=e.$$
So, $$\lim_{n\to\infty}\frac{\left(\left(1+\frac 1n\right)^n\right)^{b+\frac cn+n}}{e^n}\not=\lim_{n\to\infty}\frac{e^{b+\frac cn+n}}{e^n}.$$
What you did is like
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n=\lim_{n\to\infty}(1+0)^n=1.$$
You should know why this is impossible.
A: hint: $a_n=\left(\dfrac{\left(1+\frac{1}{n}\right)^n}{e}\right)^n \Rightarrow \ln (a_n) = \dfrac{\ln(1+\frac{1}{n})-\frac{1}{n}}{\frac{1}{n^2}}\to 0$ by L'hospitale rule with $x = \frac{1}{n} \to 0 \Rightarrow a_n \to 1$. Given that the answer is $e^{b-1/2}$, can you figure the other part ?
