calculating the limit $\lim _{n\rightarrow \infty}((4^n+3)^{1/n}-(3^n+4)^{1/n})^{n3^n}$ We need to calculating the limit 
$$
\lim _{n\rightarrow \infty}((4^n+3)^{1/n}-(3^n+4)^{1/n})^{n3^n}
$$
I have tried taking the logarithm, but the limit doesnt seem to arrive at any familiar form.
 A: I think the limit should be $e^{-12}$.
First, after you rewrite the expression within the brackets, you get 
$$
4(1+\frac{3}{4^n})^{\frac{1}{n}} - 3(1+\frac{4}{3^n})^{\frac{1}{n}}
$$
Both expressions within the two brackets can be expanded using Generalized binomial theorem, and the leading terms will be 
$$
4(1+\frac{3}{n 4^n}) - 3(1+\frac{4}{n 3^n}) + o(1) = 1+ \frac{4}{n 3^n} + O\Big(\frac{1}{n4^n}\Big)  
$$
so the limit becomes 
$$
\bigg(1 - \frac{12}{n 3^n} \bigg)^{n 3^n} \to_n e^{-12}
$$
EDIT: 
because $O\Big(\frac{1}{n4^n}\Big)$ doesn't affect the convergence rate. To see that, consider
\begin{align}
\Big(1+\frac{1}{n} +\frac{1}{n^2} \Big)^n 
&= \Big(\Big(1+\frac{1}{n}\Big)\Big(1+\frac{\frac{1}{n^2}}{\frac{n+1}{n}}\Big)\Big)^n\\
&= \Big(1+\frac{1}{n}\Big)^n\Big(1+\binom{n}{1}\frac{1}{n(n+1)} +o(1)\Big)\\
& = \Big(1+\frac{1}{n}\Big)^n\Big(1+o(1)\Big) \to_n e
\end{align}
A: Following the hint by Alex with some more detail, we have that


*

*$(4^n+3)^{1/n}=4\left(1+\frac3{4^n}\right)^\frac1n=4e^{\frac1n\log \left(1+\frac3{4^n}\right)}=4\left(1+\frac3{n4^n}+o\left(\frac1{n4^n}\right)\right)=\\=4+\frac{12}{n4^n}+o\left(\frac{12}{n4^{n}}\right)$

*$(3^n+4)^{1/n}=3\left(1+\frac4{3^n}\right)^\frac1n=3e^{\frac1n\log \left(1+\frac4{3^n}\right)}=3\left(1+\frac4{n3^n}+o\left(\frac1{n3^n}\right)\right)=\\=3+\frac{12}{n3^n}+o\left(\frac{12}{n3^{n}}\right)$


then
$$( (4^n+3)^{1/n}-(3^n+4)^{1/n} )^{n3^n}=\left(1-\frac{12}{n3^n}+o\left(\frac{12}{n3^{n}}\right)\right)^{n3^n}=$$
$$=\left[\left(1-\frac{12}{n3^n}+o\left(\frac{12}{n3^{n}}\right)\right)^{\frac{1}{-\frac{12}{n3^n}+o\left(\frac{12}{n3^{n}}\right)}}\right]^{-12+o(1)}\to e^{-12}$$
