Đáp án:f)0
g)36
h)$\frac{4}{3}$
i)2
Giải thích các bước giải:
f)$lim_{x\rightarrow 2}\frac{2-x}{\sqrt{x+7}-3}$
=$\lim_{x\rightarrow 2}\frac{-(x-2)(\sqrt{x+7}-9)}{x+7-9}$
=$\lim_{x\rightarrow 2}\frac{-(x-2)(\sqrt{x+7}-3)}{x-2}$
=$\lim_{x\rightarrow 2}-\sqrt{x+7}+3=0$
g)$\lim_{x\rightarrow 3}\frac{x^{2}-9}{\sqrt{x+1}-2}$
=$\lim_{x\rightarrow 3}\frac{(x-3)(x+3)(\sqrt{x+1}+2)}{x+1-4}$
=$\lim_{x\rightarrow 3}\frac{(x-3)(x+3)(\sqrt{x+1}+2)}{x-3}$
=$\lim_{x\rightarrow 3}(x+3)(\sqrt{x+1}+2)=36$
h)$\lim_{x\rightarrow 4}\frac{\sqrt{2x+1}-3}{\sqrt{x}-2}$
=$\lim_{x\rightarrow 4}\frac{(2x+1-9)(\sqrt{x}+2)}{(x-4)(\sqrt{2x+1}+3)}$
=$\lim_{x\rightarrow 4}\frac{(2x-8)(\sqrt{x}+2)}{(x-4)(\sqrt{2x+1}+3)}$
=$\lim_{x\rightarrow 4}\frac{2(x-4)(\sqrt{x}+2)}{(x-4)(\sqrt{2x+1}+3)}$
=$\lim_{x\rightarrow 4}\frac{2(\sqrt{x}+2)}{\sqrt{2x+1}+3}=\frac{4}{3}$
i)$\lim_{x\rightarrow -1}\frac{\sqrt{x+2}-1}{\sqrt{x+5}-2}$
=$\lim_{x\rightarrow -1}\frac{(x+2-1)(\sqrt{x+5}+2)}{(x+5-4)(\sqrt{x+2}+1)}$
=$\lim_{x\rightarrow -1}\frac{(x+1)(\sqrt{x+5}+2)}{(x+1)(\sqrt{x+2}+1)}$
=$\lim_{x\rightarrow -1}\frac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2$
