(a) Find lim _(x arrow infty) (x)/(x^3)+5 (b) lim _(x arrow-2) (x^3+2 x^2-5)/(x^2)-5 x+2

(a) To find the limit \( \lim _{x \rightarrow \infty} \frac{x}{x^{3}+5} \), we can divide both the numerator and denominator by \( x^3 \), the highest power of \( x \) in the denominator. This gives us \( \lim _{x \rightarrow \infty} \frac{1/x^2}{1+5/x^3} \). As \( x \) approaches infinity, the terms \( 1/x^2 \) and \( 5/x^3 \) approach 0, so the limit is \( \lim _{x \infty} \frac{1/x^2}{1+5/x^3} = \frac{0}{1} = 0 \).<br /><br />(b) To find the limit \( \lim _{x \rightarrow-2} \frac{x^{3}+2 x^{2}-5}{x^{2}-5 x+2} \), we can factor the numerator and denominator. The numerator factors asx+5)(x-1)(x+1) \) and the denominator factors as \( (x-2)(x-3) \). Then, we can cancel the common factor of \( (x-2) \) in the numerator and denominator. This gives us \( \lim _{x \rightarrow-2} \frac{(x+5)(x-1)(x+1)}{(x-3)} \). Now, we can substitute \( x = -2 \) into the expression to get \( \frac{(-2+5)(-2-1)(-2+1)}{-2-3} = \frac{3(-3)(-1)}{-5} = \frac{9}{5} \).