calculating the limit $\lim _{n\rightarrow \infty}((4^n+3)^{1/n}-(3^n+4)^{1/n})^{n3^n}$

Question

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.

Answer 1

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}

Answer 2

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

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

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