(D) To determine if the limit exists at $x = 1$,we calculate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$. $LHL = \lim_{x \to 1^-} f(

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$\lim_{x \to 1} f(x)$ does not exist

(D) To determine if the limit exists at $x = 1$,we calculate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$. $LHL = \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} 3(1 - h) = 3(1 - 0) = 3$. $RHL = \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} [5 - 3(1 + h)] = 5 - 3(1 + 0) = 5 - 3 = 2$. Since $LHL \neq RHL$ (i.e.,$3 \neq 2$),the limit $\lim_{x \to 1} f(x)$ does not exist.

Class 12 Mathematics Continuity and Differentiation Continuity

Medium

Let the function $f$ be defined by the equation $f(x) = \begin{cases} 3x & \text{if } 0 \le x \le 1 \\ 5 - 3x & \text{if } 1 < x \le 2 \end{cases}$,then:

A
$\lim_{x \to 1} f(x) = f(1)$
B
$\lim_{x \to 1} f(x) = 3$
C
$\lim_{x \to 1} f(x) = 2$
D
$\lim_{x \to 1} f(x)$ does not exist

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Continuity and Differentiation

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Continuity

If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

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$f(x) = \begin{cases} \frac{\log x}{x-1}, & \text{if } x \neq 1 \\ k, & \text{if } x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

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Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \ge 0 \end{cases}$. Is $f$ a continuous function?

Medium

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Match the items given in List $A$ with those of the items of List $B$:
$A$. $|x| + |x - 2|$ $I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$ $II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$ $III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$ $IV$. Continuous for all real value of $x$.
$V$. Discontinuous at all integral values of $x$.

The correct match is:

MediumTS EAMCET 2021

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Let $f(x) = \begin{cases} \operatorname{sgn}(x^2 - 3x + 2) & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$,then the number of points where $f(x)$ is continuous is (where $\operatorname{sgn}(x)$ denotes the signum function of $x$).

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$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

Let the function $f$ be defined by the equation $f(x) = \begin{cases} 3x & \text{if } 0 \le x \le 1 \\ 5 - 3x & \text{if } 1 < x \le 2 \end{cases}$,then:

Similar Questions

If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

$f(x) = \begin{cases} \frac{\log x}{x-1}, & \text{if } x \neq 1 \\ k, & \text{if } x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \ge 0 \end{cases}$. Is $f$ a continuous function?

Let $f(x) = \begin{cases} \operatorname{sgn}(x^2 - 3x + 2) & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$,then the number of points where $f(x)$ is continuous is (where $\operatorname{sgn}(x)$ denotes the signum function of $x$).

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