Let f(x) = frac{1 - x(1 + |1 - x|)}{|1 - x|} | vedclass

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

Question

(B) For $x 	o 1^{-}$, let $x = 1 - h$ where $h > 0$. Then $|1 - x| = h$.$f(1 - h) = \frac{1 - (1 - h)(1 + h)}{h} \cos \left(\frac{1}{h}ight) = \fra

Vedclass · Accepted Answer

$(A)$ and $(D)$

Vedclass · Answer

(B) For $x 	o 1^{-}$, let $x = 1 - h$ where $h > 0$. Then $|1 - x| = h$.$f(1 - h) = \frac{1 - (1 - h)(1 + h)}{h} \cos \left(\frac{1}{h}ight) = \frac{1 - (1 - h^2)}{h} \cos \left(\frac{1}{h}ight) = h \cos \left(\frac{1}{h}ight)$.Since $\lim_{h 	o 0} h \cos \left(\frac{1}{h}ight) = 0$, we have $\lim_{x 	o 1^{-}} f(x) = 0$. Thus, statement $(A)$ is true.For $x 	o 1^{+}$, let $x = 1 + h$ where $h > 0$. Then $|1 - x| = h$.$f(1 + h) = \frac{1 - (1 + h)(1 + h)}{h} \cos \left(\frac{-1}{h}ight) = \frac{1 - (1 + 2h + h^2)}{h} \cos \left(\frac{1}{h}ight) = -(2 + h) \cos \left(\frac{1}{h}ight)$.As $h 	o 0$, $-(2 + h) 	o -2$, and $\cos \left(\frac{1}{h}ight)$ oscillates between $-1$ and $1$. Thus, $\lim_{x 	o 1^{+}} f(x)$ does not exist. Thus, statement $(D)$ is true.Therefore, options $(A)$ and $(D)$ are correct.

Let $f(x) = \frac{1 - x(1 + |1 - x|)}{|1 - x|} \cos \left(\frac{1}{1 - x}\right)$ for $x \neq 1$. Which of the following statements is/are true?
$(A)$ $\lim_{x \to 1^{-}} f(x) = 0$
$(B)$ $\lim_{x \to 1^{-}} f(x) = 1$
$(C)$ $\lim_{x \to 1^{+}} f(x) = 0$
$(D)$ $\lim_{x \to 1^{+}} f(x)$ does not exist.

Similar Questions

Let $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x} + \sqrt{2}}{\sqrt{16 + \sqrt{x}}}, & x > 0 \end{cases}$
If $f(x)$ is continuous at $x = 0$,then the value of $a$ is

Discuss the continuity of the function $f$ given by $f(x) = x^{3} + x^{2} - 1$.

Let $f(x) = [x^2] \sin(\pi x)$, for $x > 0$. Then:

If $f(x) = \begin{cases} |x - 3|, & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & x < 1 \end{cases}$,then $f(x)$ is:

Find all points of discontinuity of $f,$ where $f$ is defined by $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $f(x) = \frac{1 - x(1 + |1 - x|)}{|1 - x|} \cos \left(\frac{1}{1 - x}\right)$ for $x \neq 1$. Which of the following statements is/are true?$(A)$ $\lim_{x \to 1^{-}} f(x) = 0$$(B)$ $\lim_{x \to 1^{-}} f(x) = 1$$(C)$ $\lim_{x \to 1^{+}} f(x) = 0$$(D)$ $\lim_{x \to 1^{+}} f(x)$ does not exist.