If $f: R ightarrow R$ is defined by $$ f(x)=\left\{\begin{array}{cc} \frac{\cos 3 x-\cos x}{x^2}, & ext { for } x
eq 0 \ \lambda & , ext { for } x=0 \end{array} ight. $$ and if $f$ is continuous at $x=0$, then $\lambda$ is equal to

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If $f: R ightarrow R$ is defined by $$ f(x)=\left\{\begin{array}{cc} \frac{\cos 3 x-\cos x}{x^2}, & 	ext { for } x 
eq 0 \ \lambda & , 	ext { for } x=0 \end{array}ight. $$ and if $f$ is continuous at $x=0$, then $\lambda$ is equal to

Aicharya · Accepted Answer

Given that, $ f(x)=\left\{\begin{array}{cc} \frac{\cos 3 x-\cos x}{x^2}, & 	ext { for } x 
eq 0 \ \lambda, & 	ext { for } x=0 \end{array}ight. $ Now, $ \begin{aligned} 	ext { LHL } & =\lim _{x ightarrow 0^{-}} f(x) \ & =\lim _{x ightarrow 0^{-}} \frac{\cos 3 x-\cos x}{x^2} \ & =\lim _{h ightarrow 0} \frac{\cos 3(0-h)-\cos (0-h)}{(0-h)^2} \ & =\lim _{h ightarrow 0} \frac{\cos 3 h-\cos h}{h^2} \ & =\lim _{h ightarrow 0} \frac{-3 \sin 3 h+\sin h}{2 h} \ & =\lim _{h ightarrow 0} \frac{-9 \cos 3 h+\cos h}{2} \ & =\frac{-9+1}{2}=-4 \end{aligned} $ (using L' Hospital's rule) Since, $f(x)$ is continuous at $x=0$ $ \begin{array}{ll} 	herefore & \lim _{x ightarrow 0^{-}} f(x)=f(0) \ \Rightarrow & -4=\lambda \Rightarrow \lambda=-4 \end{array} $

If $f: R \rightarrow R$ is defined by $$ f(x)=\left\{\begin{array}{cc} \frac{\cos 3 x-\cos x}{x^2}, & \text…

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