Answer:
[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]
[tex]\lim_{x\rightarrow 0 } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]
Step-by-step explanation:
[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]
[tex]= [\lim_{x\rightarrow +\infty } (x-1)]+[ \lim_{x\rightarrow +\infty } (\frac{1+ln^{2}x}{x})][/tex]
= +∞ + 0
= +∞
[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]
[tex]= [\lim_{x\rightarrow 0 } (x-1)]+[ \lim_{x\rightarrow 0 } (\frac{1+ln^{2}x}{x})][/tex]
= -1 + +∞
= +∞
