Given , f(x)={1−cos axxsinx, x≠0 12, x=0 if f is continuous at x=0, then the value of a2 must be

∵ f is continuous at x=0, thenf(0)=RHL⇒ 12=limx→0+f(x)=limh→0f(0+h)⇒ 12=limh→01−cos ahh sin h⇒ limh→01−cos ahh2(sin hh)=a22∴=a2=1

Given , f(x)={1−cos axxsinx, x≠0 12, x=0 if f is continuous at x=0, then the value of a2 must be

∵ f is continuous at x=0, thenf(0)=RHL⇒ 12=limx→0+f(x)=limh→0f(0+h)⇒ 12=limh→01−cos ahh sin h⇒ limh→01−cos ahh2(sin hh)=a22∴=a2=1

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Given , $f (x) = {\begin{matrix} \frac{1 - \cos a x}{x \sin x}, x \neq 0 \\ \frac{1}{2}, x = 0 \end{matrix}$ if $f$ is continuous at $x = 0,$ then the value of $a^{2}$ must be

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$∵ f$ is continuous at $x = 0,$ then

$f (0) = RHL$

$\Rightarrow \frac{1}{2} = \lim_{x \to 0 +} f (x) = \lim_{h \to 0} f (0 + h)$

$\Rightarrow \frac{1}{2} = \lim_{h \to 0} \frac{1 - \cos a h}{h \sin h}$

$\Rightarrow \lim_{h \to 0} \frac{1 - \cos a h}{h^{2} (\frac{\sin h}{h})} = \frac{a^{2}}{2}$

$∴ = a^{2} = 1$

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Given , f(x)={1−cos axxsinx, x≠0 12, x=0 if f is continuous at x=0, then the value of a2 must be

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Given , $f (x) = {\begin{matrix} \frac{1 - \cos a x}{x \sin x}, x \neq 0 \\ \frac{1}{2}, x = 0 \end{matrix}$ if $f$ is continuous at $x = 0,$ then the value of $a^{2}$ must be