Let f and g be real-valued functions. If lim | vedclass

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(D) Given that $\lim _{x \rightarrow 0} f(x)=1$,$\lim _{x \rightarrow 0} g(x)=\alpha$ and $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{2 / 3}

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continuous on $\left[0, \frac{\pi}{5}\right]$

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(D) Given that $\lim _{x ightarrow 0} f(x)=1$,$\lim _{x ightarrow 0} g(x)=\alpha$ and $\lim _{x ightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{2 / 3}}=\frac{7}{4}$.Substituting the limits,we get $\frac{2(1)-\alpha}{(1+7)^{2 / 3}}=\frac{7}{4}$.$\Rightarrow \frac{2-\alpha}{8^{2 / 3}}=\frac{7}{4} \Rightarrow \frac{2-\alpha}{4}=\frac{7}{4} \Rightarrow 2-\alpha=7 \Rightarrow \alpha=-5$.Now,$h(x)= \begin{cases} \sin (-5x), & 0 \leq x \leq \frac{\pi}{10} \ \cos (-10x), & \frac{\pi}{10} Since $\sin (-5x)$ and $\cos (-10x)$ are continuous functions,we only need to check continuity at $x=\frac{\pi}{10}$.$LHL = \lim _{x ightarrow \frac{\pi}{10}^-} \sin (-5x) = \sin \left(-\frac{5\pi}{10}ight) = \sin \left(-\frac{\pi}{2}ight) = -1$.$h\left(\frac{\pi}{10}ight) = \sin \left(-\frac{5\pi}{10}ight) = -1$.$RHL = \lim _{x ightarrow \frac{\pi}{10}^+} \cos (-10x) = \cos \left(-\frac{10\pi}{10}ight) = \cos (-\pi) = -1$.Since $LHL = RHL = h\left(\frac{\pi}{10}ight)$,the function $h(x)$ is continuous at $x=\frac{\pi}{10}$.Thus,$h(x)$ is continuous on $\left[0, \frac{\pi}{5}ight]$.

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