Evaluating a Simple Enough Looking Limit. I am trying to evaluate the following simple enough looking limit,
$$\lim_{x \to \infty} x^\alpha \cdot l^x, \ l\in(0,1),\ \alpha \in \mathbb{R}$$
It is not so bad when $\alpha \leq 0$, as the $l^x$ term will approach zero, but when $\alpha >0$, the limit becomes of indeterminate form, and I am not sure how to work around this. How should I proceed? I do not see how to, for instance apply the epsilon delta definition here.
A method without using the L'Hôpital rule would be extra appreciated.
 A: $$\lim_{x \to \infty} x^\alpha \cdot l^x, \ l\in(0,1),\ \alpha \in \mathbb{R}$$
Since $l\in(0,1)$ ,
$$\lim_{x \to \infty} x^\alpha \cdot \frac{1}{L^x}, \ L\in\mathbb{R^+},\ \alpha \in \mathbb{R}$$
Now observe that 
$$L^x=e^{x \cdot ln{L}}$$
$$L^x=1+\frac{x \cdot ln{L}}{1!}+\frac{(x \cdot ln{L})^2}{2!}+\frac{(x \cdot ln{L})^3}{3!}+.....+\frac{(x \cdot ln{L})^{\beta}}{\beta!}+\frac{(x \cdot ln{L})^{\beta+1}}{(\beta+1)!}.......$$
Now you can argue that for all $\alpha \in \mathbb{R^+}$ ,  there exists $\alpha \in \mathbb{N^+}$ such that $\alpha < \beta.$
$$\lim_{x \to \infty} \frac{{x^\alpha}}{1+\frac{x \cdot ln{L}}{1!}+\frac{(x \cdot ln{L})^2}{2!}+\frac{(x \cdot ln{L})^3}{3!}+.....+\frac{(x \cdot ln{L})^{\beta}}{\beta!}+ +\frac{(x \cdot ln{L})^{\beta+1}}{( \beta+1)!}.......}$$
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{\alpha}}+\frac{1 \cdot ln{L}}{{x^{\alpha-1}}1!}+\frac{(1 \cdot ln{L})^2}{{x^{\alpha-2}}2!}+\frac{(1\cdot ln{L})^3}{{x^{\alpha-3}}3!}+.....+\frac{({x^{\beta -\alpha}} \cdot ln{L}^{\beta})}{\beta!}+.......}=0$$
A: The key here is to take logs and use the standard limit $(\log x) /x\to 0$ as $x\to\infty $. By taking logs we see that it is sufficient to evaluate the limit of $$\alpha\log x+x\log l=x\left(\frac{\log x} {x} \cdot\alpha+\log l\right) $$ and the expression in parentheses tends to $\log l<0$ so that the entire expression tends to $-\infty$. It now follows by exponentiating that the desired limit is $0$.
It should also be observed that the limit does not depend on parameter $\alpha$ (well it does when $l=1$), but rather depends on $l$. For the current case $0<l<1$ the limit is $0$ and for $l>1$ the limit is $\infty$. When $l=1$ the only parameter left is $\alpha$ and the limit is easy to handle in this case.
Also $\epsilon, \delta $ definition is not used to evaluate limits, but rather used to prove limit theorems which are used to evaluate limits.
