Given f(x) = begin{cases} frac{ln(1+text{sgn} | vedclass

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

Question

(C) To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.For $x > 0$,$[x] = 0$,so $	ext{sgn}[x] = 0$.

Vedclass · Accepted Answer

For $k = 2$,$f(x)$ has a non-removable discontinuity at $x = 0$

Vedclass · Answer

(C) To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.For $x > 0$,$[x] = 0$,so $	ext{sgn}[x] = 0$. Thus,$f(x) = \frac{\ln(1+x^2)}{1-\cos x}$.$RHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(1+h^2)}{1-\cos h} = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(1+h^2)}{h^2} \cdot \frac{h^2}{1-\cos h} = 1 \cdot 2 = 2$.For $x $LHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(h^2)}{1-\cos(h)} = \mathop {\lim }\limits_{h 	o 0^+} \frac{\ln(h^2)}{h^2/2} = -\infty$.Since the $LHL$ does not exist (it is $-\infty$),the limit $\mathop {\lim }\limits_{x 	o 0} f(x)$ does not exist.Therefore,$f(x)$ has a non-removable discontinuity at $x = 0$ for any value of $k$.

Similar Questions

Let $f(x) = x^3$, $x \in [-1, 1]$. Then which of the following are correct?

If $f(x) = \begin{cases} \frac{x^4 - 16}{x - 2}, & x \neq 2 \\ 16, & x = 2 \end{cases}$,then:

If $f(x) = \begin{cases} (1 + 2x)^{1/x}, & x \ne 0 \\ e^2, & x = 0 \end{cases}$,then:

If $f(x) = |x - 2|$,then

If a function $f(x)$ defined on $[a, b]$ is discontinuous at $x=\alpha \in(a, b)$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module