Đáp án: $1.0 $ $2.2$
Giải thích các bước giải:
1.$\lim_{x\to1}\dfrac{\sqrt{x+8}-3}{x-2}$
$=\dfrac{\sqrt{1+8}-3}{1-2}$
$=0$
2.$\lim_{x\to-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}$
$=\lim_{x\to-1}\dfrac{\dfrac{x+2-1}{\sqrt{x+2}+1}}{\dfrac{x+5-2^2}{\sqrt{x+5}+2}}$
$=\lim_{x\to-1}\dfrac{\dfrac{x+1}{\sqrt{x+2}+1}}{\dfrac{x+1}{\sqrt{x+5}+2}}$
$=\lim_{x\to-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}$
$=\dfrac{\sqrt{-1+5}+2}{\sqrt{-1+2}+1}$
$=2$