$\lim _{x \rightarrow 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)+\frac{x^2-4}{x^3-2 x-4}\right]=$

$\operatorname{Lim}_{n}$ ${\rightarrow \infty} \left\{ \left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right) \left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \dots \left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right) \right\}$ is equal to

Let $f: R \rightarrow (0, \infty)$ be a strictly increasing function such that $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$. Then,the value of $\lim _{x \rightarrow \infty} \left[\frac{f(5 x)}{f(x)}-1\right]$ is equal to

If $\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}}=k$ and $\lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=l$,then $k+l=$

Let $f: (0, \infty) \rightarrow \mathbb{R}$ and $g: (0, \infty) \rightarrow \mathbb{R}$ be two functions where $g(x) = x + \frac{1}{x}$. If $1 < f(x) \cdot g(x) < 10$ for all $x > 0$,then $\lim_{x \to \infty} f(x)$ is

Let $f: R \to R$ be a positive increasing function with $\lim_{x \to \infty} \frac{f(3x)}{f(x)} = 1$. Then $\lim_{x \to \infty} \frac{f(2x)}{f(x)} = $