If f(x) = begin{cases} frac{|x-2|}{x-2}, & x | vedclass

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

Question

(D) Given $f(x) = \frac{|x-2|}{x-2}$ for $x 
eq 2$ and $f(2) = 1$. We know that $|x-2| = (x-2)$ if $x > 2$ and $|x-2| = -(x-2)$ if $x < 2$. For the r

Vedclass · Accepted Answer

$f(x)$ is discontinuous at $x=2$

Vedclass · Answer

(D) Given $f(x) = \frac{|x-2|}{x-2}$ for $x 
eq 2$ and $f(2) = 1$. We know that $|x-2| = (x-2)$ if $x > 2$ and $|x-2| = -(x-2)$ if $x For the right-hand limit: $\lim_{x ightarrow 2^{+}} f(x) = \lim_{x ightarrow 2^{+}} \frac{x-2}{x-2} = 1$. For the left-hand limit: $\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{-}} \frac{-(x-2)}{x-2} = -1$. Since $\lim_{x ightarrow 2^{+}} f(x) 
eq \lim_{x ightarrow 2^{-}} f(x)$,the limit does not exist at $x=2$. Also,$f(2) = 1$. Since the limit does not exist,the function $f(x)$ is discontinuous at $x=2$.

If $f(x) = \begin{cases} \frac{|x-2|}{x-2}, & x \neq 2 \\ 1, & x = 2 \end{cases}$,then which of the following statements is true?

Similar Questions

Which one of the following functions is not continuous on $(0, \pi )$?

Show that the function $f$ defined by $f(x) = |1 - x + |x||$,where $x$ is any real number,is a continuous function.

If $f(x) = \begin{cases} x+a, & x \leq 0 \\ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$,then $(g \circ f)(2) + (f \circ g)(-2)$ is equal to.

If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

If the function given by $f(x) = \begin{cases} -2 \sin x & -\pi \leq x < -\pi/2 \\ a \sin x + b & -\pi/2 \leq x \leq \pi/2 \\ \cos x & \pi/2 < x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,then the value of $(3a + 2b)^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module