If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

The value of $\mathop {\text{Limit}}\limits_{x \to 0} \frac{\tan(\{x\} - 1) \sin\{x\}}{\{x\}(\{x\} - 1)}$,where $\{x\}$ denotes the fractional part function,is:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $\lim _{x \rightarrow \infty} f(x)=M > 0$. Then which of the following is false?

The value of $\lim _{n}$ ${\rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]$ is

The value of $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$ is

If $f(x) = \begin{cases} \frac{\sin([x])}{[x]}, & \text{when } [x] \neq 0 \\ 0, & \text{when } [x] = 0 \end{cases}$ where $[x]$ is the greatest integer function,then $\lim_{x \to 0} f(x) = $