Tính:

Giải:

a) $$\begin{gathered} \underset{x\to \infty }{lim}\left(x^4-x^2+x-1\right) =\underset{x\to \infty }{lim}{x}^4\left(1-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}\right)=\infty \\ vì \underset{x\to \infty }{lim}{x}^4=+\infty và\underset{x\to \infty }{lim}{x}^4\left(1-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}\right)=1>0 \end{gathered}$$

b) $\underset{x\to -\infty }{lim}\left(-2x^3+3x^2-5\right)=\underset{x\to -\infty }{lim}{x}^3\left(-2+\frac{3}{x}-\frac{5}{x^2}\right)=+\infty$

Vì $\underset{x\to -\infty }{lim}{x}^3=-\infty \underset{x\to -\infty }{lim}2+\frac{3}{x}-\frac{5}{x^2}=-2<0$

c)$\underset{x\to \infty }{lim}\sqrt{x^2-2x+5}= \underset{x\to \infty }{lim}\left|x\right|\sqrt{1-\frac{2}{x}+\frac{5}{x^2}}=+\infty$

Vì $\underset{x\to \infty }{lim}\left|x\right|=+\infty và lim \sqrt{1-\frac{2}{x}+\frac{5}{x^2}}=1>0$

d)$\underset{x\to +\infty }{lim}\frac{\sqrt{x^2+1}+x}{5-2x}= \underset{x\to +\infty }{lim}\frac{x\left(\sqrt{1+{\displaystyle \frac{1}{x^2}}}+1\right)}{x\left({\displaystyle \frac{5}{x}} -2\right)}=\underset{x\to +\infty }{lim}\frac{\sqrt{1+\frac{1}{x^2}}+1}{\frac{5}{x} -2}=-1$