limit (x^3+5x)/(2x^3-x^2+4) as x->infinity(b) $\begin{aligned}[t] \lim\limits_{x\to\infty} \bigl( \sqrt{9x^2+x}-3x \bigr) &= \lim\limits_{x\to\infty} \bigl( \sqrt{9x^2+x}-3x \bigr)\dfrac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x} \\ &= \lim\limits_{x\to\infty} \dfrac{x}{\sqrt{9x^2+x}+3x} = \lim\limits_{x\to\infty} \dfrac{1}{\sqrt{9+\frac{1}{x}}+3} = \dfrac{1}{6} \end{aligned}$
limit sqrt(9x^2+x)-3x as x->infinity(c) $\lim\limits_{x\to\infty} (x-\sqrt{x}) = \lim\limits_{x\to\infty} \sqrt{x}(\sqrt{x}-1) = \lim\limits_{x\to\infty} \sqrt{x} \cdot \lim\limits_{x\to\infty} (\sqrt{x}-1) = \infty$
limit x-sqrt(x) as x->infinity(d) $\lim\limits_{x\to-\infty} (x^4+x^5) = \lim\limits_{x\to-\infty} x^4(1+x) = -\infty$
limit x^4+x^5 as x->-infinity