# Limit $\lim_{t \to{0^+}}{t^{N-1}\left[\ln \left(\ln\left( 1+\frac{1}{t} \right) \right) \right]^N }=0$

I have found the this exercise in my book of introduction to calculus, could someone help me with it?

Let $$N\geq 2$$ be a natural number, prove that the following limit is equal to $$0$$: $$\displaystyle\lim_{t \to{0^+}}{t^{N-1}\left[\ln \left(\ln\left( 1+\frac{1}{t} \right) \right) \right]^N }.$$

I tried to use L’Hospital's rule but the successive derivatives has a very complicated expression. So, is there an easier way to solve it? Any help will be appreciated.

• I am really sorry. I forgot the N-power of the second factor. I have edited it May 10, 2020 at 14:44
• It is outside. I have edited it, I hope it is clearer now May 10, 2020 at 14:56

Substitute $$u=\ln\left(1+\frac 1t\right)$$ to get
$$\lim_{u\to \infty} \frac{\ln(u)}{(e^u-1)^{N-1}}$$
$$=\lim_{u\to \infty} \frac{1}{u\cdot (N-1)\cdot (e^u-1)^{N-2} \cdot e^u} =0$$
$$\lim_{t \to{0^+}}{t^{N-1}\ln \left(\ln\left( 1+\frac{1}{t}\right) \right)^N } =\lim_{s \to{+\infty}}\frac{\ln \left(\ln\left( 1+s \right) \right)^N}{s^{N-1} } \leq \lim_{s \to{+\infty}}s\left(\frac{\ln\ln( 1+s ) }{s^1}\right)^N = \lim_{s \to{+\infty}}\left(\frac{\ln\ln( 1+s ) }{s^{(1-\frac{1} {N})}}\right)^N = 0$$ as the (broken) polynomial in the denominator still beats the logarithm in the enumerator.