$\lim_{x \to 0} \left[ \frac{x \cdot \log(1 + 4x)}{(e^{4x} - 1)^2} \right] = \dots$

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^n}}} \right)$ equals

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x),$ where $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$

Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$,where $x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim_{x \to \alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ is equal to:

$\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3} = $

$\lim _{n \rightarrow \infty} P\left(1+\frac{r}{100 n}\right)^{t n} =$