For each t in mathbb{R},let [t] be the greate | vedclass

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

Question

(B) We are evaluating the limit $\lim_{x 	o 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$.As $x 	o 1^+$,we have $x >

Vedclass · Accepted Answer

equals $0$

Vedclass · Answer

(B) We are evaluating the limit $\lim_{x 	o 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$.As $x 	o 1^+$,we have $x > 1$,so $|x| = x$ and $|1 - x| = x - 1$.Also,for $x$ slightly greater than $1$,$1 - x$ is slightly less than $0$,so $[1 - x] = -1$.Substituting these into the expression:$\lim_{x 	o 1^+} \frac{(1 - x + \sin(x - 1)) \sin(-\frac{\pi}{2})}{(x - 1)^2}$Since $\sin(-\frac{\pi}{2}) = -1$,the expression becomes:$\lim_{x 	o 1^+} \frac{-(1 - x + \sin(x - 1))}{(x - 1)^2} = \lim_{x 	o 1^+} \frac{(x - 1) - \sin(x - 1)}{(x - 1)^2}$Let $h = x - 1$. As $x 	o 1^+$,$h 	o 0^+$. The limit becomes:$\lim_{h 	o 0^+} \frac{h - \sin(h)}{h^2}$Using the Taylor series expansion $\sin(h) = h - \frac{h^3}{6} + \dots$:$\lim_{h 	o 0^+} \frac{h - (h - \frac{h^3}{6} + \dots)}{h^2} = \lim_{h 	o 0^+} \frac{\frac{h^3}{6}}{h^2} = \lim_{h 	o 0^+} \frac{h}{6} = 0$.

For each $t \in \mathbb{R}$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ is:

Similar Questions

If $a, b, c$ and $d$ are positive,then $\lim_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx}$ is equal to:

$\lim _{x \rightarrow \infty}\left(\frac{x+5}{x+2}\right)^{x+3}$ equals

$\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

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

Let $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$ and $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$,where $[ \cdot ]$ denotes the greatest integer function. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module