If $f(x) = \begin{cases} \frac{\sin(1+[x])}{[x]}, & \text{for } [x] \neq 0 \\ 0, & \text{for } [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer function,then $\lim_{x \rightarrow 0^{-}} f(x)$ is equal to
- A$-1$
- B$0$
- C$\sin(1)$
- D$1$
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$\lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right]$ is equal to
Consider the following statements:
Statement $1$: $\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = 1$ (where $a+b+c \neq 0$).
Statement $2$: $\lim _{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \frac{1}{4}$.
Statement $1$: $\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = 1$ (where $a+b+c \neq 0$).
Statement $2$: $\lim _{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \frac{1}{4}$.
MediumKCET 2021
View Solution$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ is
Medium
View SolutionIf $\mathop {\lim }\limits_{x \to 0} \phi (x) = {a^3}, (a \ne 0)$; then $\mathop {\lim }\limits_{x \to 0} \phi \left( {\frac{x}{a}} \right)$ is equal to :-
If $\mathop {\lim }\limits_{x \to \infty } \frac{e^{\mu x} + 5}{e^{100x} + 7}$ exists,then the sum of all possible positive integral values of $\mu$ is:
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