Answer
a) $\infty$
b) $-5$
Work Step by Step
a) When $x\rightarrow -\infty$ we have:
$\displaystyle\lim_{x\rightarrow -\infty}f(x)=\displaystyle\lim_{x\rightarrow -\infty}(2x^2+5)=+\infty$
b) When $x\rightarrow \infty$ we have:
$\displaystyle\lim_{x\rightarrow \infty}f(x)=\displaystyle\lim_{x\rightarrow \infty}\dfrac{3-5x^3}{1+4x+x^3}$.
We need to divide both numerator and denominator of $\frac{3 - 5x^3}{1 + 4x + x^3}$ by $x^3$. We get
$\displaystyle\lim_{x\rightarrow \infty}\dfrac{3-5x^3}{1+4x+x^3}=\displaystyle\lim_{x\rightarrow \infty}\frac{\frac{3}{x^3} - 5}{\frac{1}{x^3} + \frac{4}{x^2} + 1}$
$=\dfrac{0-5}{0+0+1}=-5$.
