For any $\lambda, \theta, \gamma>0$ show that $\lim_{n \to \infty}n^\gamma \exp\left[-\frac{n^\lambda}{\ln(n)^\theta}\right]=0$
By L'Hopital's rule, taking $k =\lceil \theta \rceil$, we know that $$\lim_{n \to \infty}\frac{n^\lambda}{\ln(n)^\theta} = \lim_{n \to \infty} \frac{\lambda n^{\lambda}}{\theta \ln(n)^{\theta-1}} = \cdots = \lim_{n \to \infty} \left[\frac{\lambda(\lambda-1)\cdots (\lambda-k+1)}{\theta(\theta-1)\cdots (\theta-k+1)}\right]\frac{n^\lambda}{\ln(n)^{\theta-k}} = \infty.$$
Now, observe that $$\ln\left(n^\gamma \exp\left[-\frac{n^\lambda}{\ln(n)^\theta}\right]\right) = \gamma \ln(n)-\frac{n^\lambda}{\ln(n)^\theta} = \frac{\gamma \ln(n)^{\theta+1}-n^\lambda}{\ln(n)^\theta}.$$ $$\lim_{n \to \infty} \frac{\gamma \ln(n)^{\theta+1}-n^\lambda}{\ln(n)^\theta} = \lim_{n\to \infty}\frac{\gamma (\theta+1)\frac{\ln(n)^\theta}{n}-\lambda n^{\lambda-1}}{\theta \frac{\ln(n)^{\theta-1}}{n}} = \lim_{n \to \infty}\frac{\gamma(\theta+1)\ln(n)^\theta-\lambda n^\lambda}{\theta \ln(n)^{\theta-1}}=\cdots$$ $$\cdots =\lim_{n \to \infty} \left[\frac{\gamma(\theta+1)\theta(\theta-1)\cdots (\theta-k+1)}{\theta (\theta-1)\cdots (\theta-k)}\right]\frac{\ln(n)^{\theta-k}}{\ln(n)^{\theta-k-1}}-\frac{\lambda^kn^\lambda}{\theta\cdots (\theta-k)\ln(n)^{\theta-k-1}}=-\infty.$$
Which allows us to say that, for every $M\in \mathbb{R}$ exists $N_M$ such that $$\ln\left(n^\gamma \exp\left[-\frac{n^\lambda}{\ln(n)^\theta}\right]\right)<-M\; \forall n>N_M.$$
So $$0\leq n^\gamma \exp\left[-\frac{n^\lambda}{\ln(n)^\theta}\right]<\exp[-M]\; \forall n>N_M,$$ which implies that $\lim_{n \to \infty}n^\gamma \exp\left[-\frac{n^\lambda}{\ln(n)^\theta}\right]=0$
Is this correct? And if so, are there cleanest ways to show this statement?
For example, one could find a function $g:\mathbb{R}\to \mathbb{R}$ with $\lim_{n \to \infty}g(n)=0$ such that $$n^\gamma e^{-\tfrac{n^\lambda}{\ln(n)^\theta}}<g(n)\;\forall n.$$
I couldn' think of any such functions.
Let $$f(n;\gamma,\lambda,\theta) = n^\gamma \exp \left(-\frac{n^\lambda}{(\log n)^\theta}\right) = \exp\left( \gamma \log n \right) \exp \left(-\frac{e^{\lambda \log n}}{(\log n)^\theta}\right)= \exp \left( \gamma \log n - \frac{e^{\lambda \log n}}{(\log n)^\theta}\right).$$ This suggests defining $$g(x;\gamma,\lambda,\theta) = \gamma x - \frac{e^{\lambda x}}{x^\theta}$$ and considering the limit of $g$ as $x \to \infty$. It is not too difficult to show $g \to -\infty$ as long as the given conditions on $\gamma, \lambda, \theta$ are satisfied.