Let f(x) = lim_{y to 0} frac{(1 - cos(xy))tan | vedclass

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(C) To evaluate the limit $f(x) = \lim_{y 	o 0} \frac{(1 - \cos(xy))	an(xy)}{y^3}$,we multiply and divide by $(xy)^3$:$f(x) = \lim_{y 	o 0} \left(

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$3$

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(C) To evaluate the limit $f(x) = \lim_{y 	o 0} \frac{(1 - \cos(xy))	an(xy)}{y^3}$,we multiply and divide by $(xy)^3$:$f(x) = \lim_{y 	o 0} \left( \frac{1 - \cos(xy)}{(xy)^2} \cdot \frac{	an(xy)}{xy} \cdot \frac{x^3 y^3}{y^3} ight)$Using standard limits $\lim_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$ and $\lim_{	heta 	o 0} \frac{	an 	heta}{	heta} = 1$,we get:$f(x) = \frac{1}{2} \cdot 1 \cdot x^3 = \frac{x^3}{2}$.Now,we need to find the number of solutions for the equation $f(x) = \sin x$,which is $\frac{x^3}{2} = \sin x$,or $x^3 = 2 \sin x$.Let $g(x) = x^3 - 2 \sin x$. We look for roots where $g(x) = 0$.At $x = 0$,$g(0) = 0 - 2(0) = 0$. So,$x = 0$ is a solution.For $x > 0$,$x^3 = 2 \sin x$ has one positive solution because $x^3$ is strictly increasing and $2 \sin x$ is bounded by $2$.For $x  0$. Then $(-t)^3 = 2 \sin(-t) \implies -t^3 = -2 \sin t \implies t^3 = 2 \sin t$. This gives one negative solution.Thus,there are $3$ solutions in total: $x = 0$,$x \approx 1.41$,and $x \approx -1.41$.

Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x, x \in R$ is:

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