Chủ đề này có 2 trả lời

[chidungdijiyeon]

Cho giới hạn L = $\lim_{x\rightarrow 0}\frac{sinx - tanx}{xe^{ax} - ln(1+x)}$. Tìm a để L$\neq 0$

[Lee LOng]

$\frac{sin{x}-tan{x}}{xe^{ax}-ln(1+x)}=\frac{x-\frac{x^3}{3!}-\frac{x-\frac{x^3}{3!}+o(x^3)}{1-\frac{x^2}{2}}}{x(1+x+\frac{x^2}{2})^{a}-ln(1+x)+o(x^3)}=\frac{x-\frac{x^3}{3!}-(x+\frac{x^3}{3})}{x[1+a(x+\frac{x^2}{2})+\frac{a(a-1)}{2}](x+\frac{x^2}{2})^{2}-(x-\frac{x^2}{2}+\frac{x^3}{3})+o(x^3)}=\frac{-x^3}{x^2(2a+1)+x^3(a^2-\frac{2}{3})}\Rightarrow a=\frac{-1}{2}$ 

[chidungdijiyeon]

$\frac{sin{x}-tan{x}}{xe^{ax}-ln(1+x)}=\frac{x-\frac{x^3}{3!}-\frac{x-\frac{x^3}{3!}+o(x^3)}{1-\frac{x^2}{2}}}{x(1+x+\frac{x^2}{2})^{a}-ln(1+x)+o(x^3)}=\frac{x-\frac{x^3}{3!}-(x+\frac{x^3}{3})}{x[1+a(x+\frac{x^2}{2})+\frac{a(a-1)}{2}](x+\frac{x^2}{2})^{2}-(x-\frac{x^2}{2}+\frac{x^3}{3})+o(x^3)}=\frac{-x^3}{x^2(2a+1)+x^3(a^2-\frac{2}{3})}\Rightarrow a=\frac{-1}{2}$

cảm ơn bạn nha