Let f: R rightarrow R be defined by f(x)=begi | vedclass

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

Question

(B) Given that $f(x)$ is continuous at $x=0$,therefore $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = 2$. First,find the Left

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given that $f(x)$ is continuous at $x=0$,therefore $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = 2$. First,find the Left Hand Limit $(LHL)$: $\lim_{x \rightarrow 0^-} f(x) = \lim_{h \rightarrow 0} f(-h) = \lim_{h \rightarrow 0} \left( \beta + \left[ \frac{\sin(-h) - (-h)}{(-h)^3} \right] \right) = \lim_{h \rightarrow 0} \left( \beta + \left[ \frac{-\sin h + h}{-h^3} \right] \right) = \lim_{h \rightarrow 0} \left( \beta + \left[ \frac{\sin h - h}{h^3} \right] \right)$. Using the Taylor series expansion $\sin h = h - \frac{h^3}{6} + \dots$,we get $\frac{\sin h - h}{h^3} = \frac{-h^3/6}{h^3} = -\frac{1}{6}$. Thus,$LHL$ $= \beta + [-1/6] = \beta - 1$. Since $LHL$ $= f(0)$,we have $\beta - 1 = 2 \Rightarrow \beta = 3$. Next,find the Right Hand Limit $(RHL)$: $\lim_{x \rightarrow 0^+} f(x) = \lim_{h \rightarrow 0} \left( \alpha + \frac{\sin [h]}{h} \right)$. For $0 Since $RHL$ $= f(0)$,we have $\alpha = 2$. Finally,$\beta - \alpha = 3 - 2 = 1$.

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} 2 \sin \left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^2 + x + b|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{cases}$. If $f(x)$ is continuous on $R$,then $a + b$ equals ..... .

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to:

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then

Prove that every rational function is continuous at every point in its domain.

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module