Giải Bài tập 5.28 trang 124 sgk Toán 11 tập 1 Kết nối

a) $\underset{x\rightarrow 7}{lim}\frac{\sqrt{x+2}-3}{x-7}=\underset{x\rightarrow 7}{lim}\frac{1}{\sqrt{x+2}+3}=\frac{1}{6}$

b) $\underset{x\rightarrow 1}{lim}\frac{x^{3}-1}{x^{2}-1}=\underset{x\rightarrow 1}{lim}\frac{x^{2}+x+1}{x+1}=\frac{3}{2}$

c)$\underset{x\rightarrow 1}{lim}\frac{2-x}{(1-x)^{2}}=\underset{x\rightarrow 1}{lim}[(2-x)(\frac{1}{(1-x)^{2}})]$

$\underset{x\rightarrow 1}{lim}(2-x)=1$

$\underset{x\rightarrow 1}{lim}(\frac{1}{(1-x)^{2}})=+\infty $

$\Rightarrow \underset{x\rightarrow 1}{lim}\frac{2-x}{(1-x)^{2})}=+\infty $

d) $\underset{x\rightarrow -\infty }{lim}\frac{x+2}{\sqrt{4x^{2}+1}}=\underset{x\rightarrow -\infty }{lim}\frac{1+\frac{2}{x}}{-\sqrt{4+\frac{1}{x^{2}}}}=-\frac{1}{2}$