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

The value of $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)$ is

The function $f(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$ is identical to which of the following functions?

If $\lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)] = 33 \cdot \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot [1^k + 2^k + 3^k + \ldots + n^k]$,then the integral value of $k$ is equal to $....$

Find $\mathop {\lim }\limits_{x \to 5} f(x),$ where $f(x)=|x|-5$.

Let $f(x) = \frac{x \cdot 2^x - x}{1 - \cos x}$ and $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} \right)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module