[x] represents the greatest integer function. | vedclass

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(A) Given the limit $\lim _{x \rightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$. Since $x \rightarrow 0^{+}$,we have $[x] = 0$. Substituting t

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$\sqrt{10}$

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(A) Given the limit $\lim _{x ightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$. Since $x ightarrow 0^{+}$,we have $[x] = 0$. Substituting this into the expression: $\lim _{x ightarrow 0^{+}} \frac{\cos(0) - \cos(kx - 0)}{x^2} = 5$ $\lim _{x ightarrow 0^{+}} \frac{1 - \cos(kx)}{x^2} = 5$ Using the limit formula $\lim _{	heta ightarrow 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,we rewrite the expression: $\lim _{x ightarrow 0^{+}} \frac{1 - \cos(kx)}{(kx)^2} \cdot k^2 = 5$ $\frac{1}{2} \cdot k^2 = 5$ $k^2 = 10$ $k = \sqrt{10}$.

$[x]$ represents the greatest integer function. If $\lim _{x \rightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$,then $k=$

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