Cho a,b là các số thực khác 0. Tìm lim [{1 – căn(ax+1)} / {sin(bx)}] x–>0 25/10/2021 Bởi Caroline Cho a,b là các số thực khác 0. Tìm lim [{1 – căn(ax+1)} / {sin(bx)}] x–>0
$\lim\limits_{x\to 0}\dfrac{1-\sqrt{ax+1} }{\sin bx}$ $=\lim\limits_{x\to 0}\dfrac{1-ax-1}{\sin bx(1+\sqrt{ax+1})}$ $=\lim\limits_{x\to 0}\dfrac{-ax}{\sin bx(1+\sqrt{ax+1})}$ $=\lim\limits_{x\to 0}\dfrac{bx}{\sin bx}.\dfrac{-a}{b(1+\sqrt{ax+1})}$ $=1.\dfrac{-a}{b(1+1)}$ $=\dfrac{-a}{2b}$ Bình luận
Giải thích các bước giải: Ta có: $\lim_{x\to0}\dfrac{1-\sqrt{ax+1}}{\sin(bx)}$ $=\lim_{x\to0}\dfrac{\dfrac{1-(ax+1)}{1+\sqrt{ax+1}}}{\sin(bx)}$ $=\lim_{x\to0}\dfrac{\dfrac{-ax}{1+\sqrt{ax+1}}}{\sin(bx)}$ $=\lim_{x\to0}\dfrac{\dfrac{-a}{1+\sqrt{ax+1}}}{\dfrac{\sin(bx)}{x}}$ $=\lim_{x\to0}\dfrac{\dfrac{-a}{b(1+\sqrt{ax+1})}}{\dfrac{\sin(bx)}{bx}}$ $=\dfrac{\dfrac{-a}{b(1+\sqrt{a\cdot 0+1})}}{1}$ $=-\dfrac{a}{2b}$ Bình luận
$\lim\limits_{x\to 0}\dfrac{1-\sqrt{ax+1} }{\sin bx}$
$=\lim\limits_{x\to 0}\dfrac{1-ax-1}{\sin bx(1+\sqrt{ax+1})}$
$=\lim\limits_{x\to 0}\dfrac{-ax}{\sin bx(1+\sqrt{ax+1})}$
$=\lim\limits_{x\to 0}\dfrac{bx}{\sin bx}.\dfrac{-a}{b(1+\sqrt{ax+1})}$
$=1.\dfrac{-a}{b(1+1)}$
$=\dfrac{-a}{2b}$
Giải thích các bước giải:
Ta có:
$\lim_{x\to0}\dfrac{1-\sqrt{ax+1}}{\sin(bx)}$
$=\lim_{x\to0}\dfrac{\dfrac{1-(ax+1)}{1+\sqrt{ax+1}}}{\sin(bx)}$
$=\lim_{x\to0}\dfrac{\dfrac{-ax}{1+\sqrt{ax+1}}}{\sin(bx)}$
$=\lim_{x\to0}\dfrac{\dfrac{-a}{1+\sqrt{ax+1}}}{\dfrac{\sin(bx)}{x}}$
$=\lim_{x\to0}\dfrac{\dfrac{-a}{b(1+\sqrt{ax+1})}}{\dfrac{\sin(bx)}{bx}}$
$=\dfrac{\dfrac{-a}{b(1+\sqrt{a\cdot 0+1})}}{1}$
$=-\dfrac{a}{2b}$