lim _(xarrow infty )((2x-3)/(2x+1))^5x;lim _(xarrow +2)frac (1-sin^3x){

Let's correct the explanation for the second limit:<br /><br />For the second limit, $\lim_{x \to +2} \frac{1 - \sin^3 x}{x - 2}$, we can use L'Hôpital's Rule since it is in the form of $\frac{0}{0}$.<br /><br />Taking the derivatives of the numerator and denominator:<br /><br />$\frac{d}{dx}(1 - \sin^3 x) = -3\sin^2 x \cdot \cos x$<br /><br />$\frac{d}{dx}(x - 2) = 1$<br /><br />Now, applying L'Hôpital's Rule:<br /><br />$\lim_{x \to +2} \frac{1 - \sin^3 x}{x - 2} = \lim_{x \to +2} \frac{-3\sin^2 x \cdot \cos x}{1} = -3\sin^2(2)\cos(2) = -3 \cdot 1 \cdot 1 = -3$<br /><br />So, the correct answer for the second limit is $-3$.