Consider the function f(x) = begin{cases} fra | vedclass

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

Question

(B) To check the continuity at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-hand limit

Vedclass · Accepted Answer

has an isolated point removable discontinuity

Vedclass · Answer

(B) To check the continuity at $x = 2$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 2$.$1$. Left-hand limit $(LHL)$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} \frac{x}{[x]}$. Since $1 \leqslant x $2$. Right-hand limit $(RHL)$: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} \sqrt{6-x} = \sqrt{6-2} = \sqrt{4} = 2$.$3$. Value of the function: $f(2) = 1$.Since $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = 2$,but $f(2) = 1$,the limit exists but is not equal to the function value. This type of discontinuity,where the limit exists but the point is defined elsewhere,is called a removable discontinuity (specifically,an isolated point removable discontinuity).

Consider the function $f(x) = \begin{cases} \frac{x}{[x]} & \text{if } 1 \leqslant x < 2 \\ 1 & \text{if } x = 2 \\ \sqrt{6-x} & \text{if } 2 < x \leqslant 3 \end{cases}$ where $[x]$ denotes the greatest integer function. At $x = 2$,the function:

Similar Questions

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{\sin x - \sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$ is continuous on $R$,then $f(0) = $

Let $[x]$ represent the greatest integer not more than $x$. The discontinuous points of the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$ lie in the interval

If $f(x) = \begin{cases} -x^2, & \text{when } x \le 0 \\ 5x - 4, & \text{when } 0 < x \le 1 \\ 4x^2 - 3x, & \text{when } 1 < x < 2 \\ 3x + 4, & \text{when } x \ge 2 \end{cases}$,then:

Examine the following function for continuity: $f(x) = x - 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module