If lim_{x to k} frac{x^3 - k^3}{x^2 - k^2} = | vedclass

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(A) Step $1$: Evaluate the left-hand limit $(LHL)$.$\lim_{x 	o k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x 	o k} \frac{(x - k)(x^2 + xk + k^2)}{(x - k)

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$\frac{4}{3}$

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(A) Step $1$: Evaluate the left-hand limit $(LHL)$.$\lim_{x 	o k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x 	o k} \frac{(x - k)(x^2 + xk + k^2)}{(x - k)(x + k)} = \lim_{x 	o k} \frac{x^2 + xk + k^2}{x + k} = \frac{k^2 + k^2 + k^2}{k + k} = \frac{3k^2}{2k} = \frac{3k}{2}$.Step $2$: Evaluate the right-hand limit $(RHL)$.$\lim_{x 	o 0} \frac{1 - \cos(2x)}{x \sin x} = \lim_{x 	o 0} \frac{2 \sin^2 x}{x \sin x} = \lim_{x 	o 0} \frac{2 \sin x}{x} = 2(1) = 2$.Step $3$: Equate $LHL$ and $RHL$ to find $k$.$\frac{3k}{2} = 2 \implies 3k = 4 \implies k = \frac{4}{3}$.

If $\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to 0} \frac{1 - \cos(2x)}{x \sin x}$, then the value of $k$ is...

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