mathop {Lim}limits_{x to {0^ - }} sin^{-1}([t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) As $x 	o 0^-$,$	an x$ approaches $0$ from the negative side,i.e.,$-1 < 	an x < 0$. Therefore,$[	an x] = -1$. Substituting this into the limit:

Vedclass · Accepted Answer

$2 - \frac{\pi}{2}$

Vedclass · Answer

(D) As $x 	o 0^-$,$	an x$ approaches $0$ from the negative side,i.e.,$-1 Therefore,$[	an x] = -1$. Substituting this into the limit: $\mathop {Lim}\limits_{x 	o {0^ - }} \sin^{-1}([	an x]) = \sin^{-1}(-1) = -\frac{\pi}{2}$. So,$l = -\frac{\pi}{2}$. We need to find $\{l\} = \{-\frac{\pi}{2}\}$. Using the property $\{x\} = x - [x]$,we have $\{ -\frac{\pi}{2} \} = -\frac{\pi}{2} - [-\frac{\pi}{2}]$. Since $-1.57 Thus,$\{ -\frac{\pi}{2} \} = -\frac{\pi}{2} - (-2) = 2 - \frac{\pi}{2}$.

$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x])$ $= l$,then $\{l\}$ is equal to,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer and fractional part functions,respectively.

Similar Questions

$\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$ is equal to

If $\lim_{x}$ ${\rightarrow 0} \left\{ \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4} \right) \right\} = 2^{-k}$,then the value of $k$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$ is equal to

Let for all $x > 0$, $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$, then

If $f: R \rightarrow R$ is defined by $f(x) = [x-3] + |x-4|$ for $x \in R$,then $\lim_{x \rightarrow 3^{-}} f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module