Solutions

i)

$$\begin{align*} \lim_{x\to\infty}\frac{x+\sin^3 x}{5x+6} &=\lim_{x\to\infty}\left(\frac{x}{5x+6}+\frac{\sin^3 x}{5x+6}\right) \\ &=\lim_{x\to\infty}\frac{1}{5+6x^{-1}}+\frac{\sin^3 x}{5x+6} \\ &=0+0 \\ &=0. \end{align*}$$

ii)

$$\lim_{x\to\infty}\frac{x\sin x}{x^2+5} =\lim_{x\to\infty}\frac{\sin x}{x+5x^{-1}}=0.$$

iii)

$$\begin{align*} \lim_{x\to\infty}\left(\sqrt{x^2+x}-x\right) &=\lim_{x\to\infty}\left(\sqrt{x^2(1+x^{-1})}-x\right) \\ &=\lim_{x\to\infty}\left(x\sqrt{1+x^{-1}}-x\right) \\ &=0. \end{align*}$$

iv)

$$\lim_{x\to\infty}\frac{x^2(1+\sin^2x)}{(x+\sin x)^2} =\lim_{x\to\infty}\frac{1+\sin^2x}{1+2x^{-1}\sin x+x^{-2}\sin^2x}.$$

Hence the limit does not exist because

$$\lim_{x\to\infty}\frac{1}{1+2x^{-1}\sin x+x^{-2}\sin^2x}=1,$$

but

$$\lim_{x\to\infty}\frac{\sin^2x}{1+2x^{-1}\sin x+x^{-2}\sin^2x}$$

does not exist.