How to find $\lim_{n\to ∞} \Big[ \Big(1+ \frac1n\Big)^n - \Big(1+\frac1n\Big) \Big]^{-n }$?

Question

Could someone give me a hint on how to calculate this limit?

$$\lim_{n\to ∞} \Big[ \Big(1+ \frac1n\Big)^n - \Big(1+\frac1n\Big) \Big]^{-n }$$

I tried taking the logarithm, but after that what should I do with the the $-n$ that tends to $-\infty$.

Answer 1

You can write (1+1/n)^n as e due to standard limits. So the question is essentially (e-1-1/n)^-n. If you put n as -inf, you get (e-1)^-inf. any finite non zero positive integer raised to -inf is zero. Hence the answer is zero

Answer 2

Just as Did commented, consider $$A=\Big(1+ \frac1n\Big)^n - \Big(1+\frac1n\Big)$$ and apply the binomial theorem. You the have $$A=(e-1)-\frac{1+\frac{e}{2}}{n}+\frac{11 e}{24 n^2}+\cdots$$ Making the long division $$\frac 1A=\frac{1}{e-1}+\frac{2+e}{2 (e-1)^2 }\frac 1n+\cdots$$ and you are looking for the limit of $\frac 1{A^n}$.