If f(x) = frac{5x operatorname{cosec}(sqrt{x} | vedclass

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(C) Given $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$.Substituting $x^2$ for $x$,we get:$f(x^2) = \f

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$5$

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(C) Given $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$.Substituting $x^2$ for $x$,we get:$f(x^2) = \frac{5x^2 \operatorname{cosec}(x) - 1}{(x^2 - 2) \operatorname{cosec}(x)} = \frac{5x^2 \operatorname{cosec}(x)}{(x^2 - 2) \operatorname{cosec}(x)} - \frac{1}{(x^2 - 2) \operatorname{cosec}(x)}$.$f(x^2) = \frac{5x^2}{x^2 - 2} - \frac{\sin(x)}{x^2 - 2}$.Now,taking the limit as $x \rightarrow \infty$:$\lim_{x \rightarrow \infty} f(x^2) = \lim_{x \rightarrow \infty} \left( \frac{5x^2}{x^2 - 2} - \frac{\sin(x)}{x^2 - 2} \right)$.Since $\lim_{x \rightarrow \infty} \frac{5x^2}{x^2 - 2} = 5$ and $\lim_{x \rightarrow \infty} \frac{\sin(x)}{x^2 - 2} = 0$ (as $-1 \leq \sin(x) \leq 1$ and $x^2 - 2 \rightarrow \infty$),$\lim_{x \rightarrow \infty} f(x^2) = 5 - 0 = 5$.

If $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$,then $\lim_{x \rightarrow \infty} f(x^2) = $

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