Let f:R→R be defined by f(x)=α+sin[x]x,ifx>02,ifx=0β+sinx−xx3,ifx<0 where [x] denotes the integral part of x. If f continuous at x = 0, then β-α=

LHLlimx→0 β+sinx-xx3⇒use l hospital rule =β+limx→0 cosx-13x2=β+limx→0 -sinx6x=β-1RHL=α+0x=αα+0=2=β−1α=β−1⇒β−α=1

Let f:R→R be defined by f(x)=α+sin[x]x,ifx>02,ifx=0β+sinx−xx3,ifx<0 where [x] denotes the integral part of x. If f continuous at x = 0, then β-α=

LHLlimx→0 β+sinx-xx3⇒use l hospital rule =β+limx→0 cosx-13x2=β+limx→0 -sinx6x=β-1RHL=α+0x=αα+0=2=β−1α=β−1⇒β−α=1

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Let $f : R \to R$ be defined by $f (x) = \{\begin{cases} \begin{matrix} α + \frac{\sin [x]}{x} & , i f & x > 0 \\ 2 & , i f & x = 0 \\ β + [\frac{\sin x - x}{x^{3}}] & , i f & x < 0 \end{matrix} \end{cases}$ where [x] denotes the integral part of x. If f continuous at x = 0, then $β - α =$

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LHL $\lim_{x \to 0} β + [\frac{\sin x - x}{x^{3}}] \Rightarrow u s e l h o s p i t a l r u l e = β + \lim_{x \to 0} [\frac{\cos x - 1}{3 x^{2}}] = β + \lim_{x \to 0} [\frac{- \sin x}{6 x}] = β - 1$

RHL= $α + \frac{0}{x} = α$

$\begin{array}{l} α + 0 = 2 = β - 1 \\ α = β - 1 \Rightarrow β - α = 1 \end{array}$

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Let f:R→R be defined by f(x)=α+sin[x]x,ifx>02,ifx=0β+sinx−xx3,ifx<0 where [x] denotes the integral part of x. If f continuous at x = 0, then β-α=

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Let $f : R \to R$ be defined by $f (x) = \{\begin{cases} \begin{matrix} α + \frac{\sin [x]}{x} & , i f & x > 0 \\ 2 & , i f & x = 0 \\ β + [\frac{\sin x - x}{x^{3}}] & , i f & x < 0 \end{matrix} \end{cases}$ where [x] denotes the integral part of x. If f continuous at x = 0, then $β - α =$