複雜函數求極限問題？

01-05

倒代換+拉格朗日中值定理即可

直接用小量近似看出來答案是2吧，反正是填空題…

$[egin{gathered} mathop {lim }limits_{x o infty } xleft[ {sin ln (1 + frac{3}{x}) - sin ln (1 + frac{1}{x})} ight] hfill \ = mathop {lim }limits_{x o infty } xleft[ {2cos frac{{ln (1 + frac{3}{x}) + ln (1 + frac{1}{x})}}{2}sin frac{{ln (1 + frac{3}{x}) - ln (1 + frac{1}{x})}}{2}} ight] hfill \ = 2mathop {lim }limits_{x o infty } xleft[ {cos frac{{ln (1 + frac{3}{x})(1 + frac{1}{x})}}{2}sin frac{{ln frac{{(1 + frac{3}{x})}}{{(1 + frac{1}{x})}}}}{2}} ight] hfill \ = 2mathop {lim }limits_{x o infty } left[ {frac{{sin frac{{ln frac{{(1 + frac{3}{x})}}{{(1 + frac{1}{x})}}}}{2}}}{{frac{1}{x}}}} ight] hfill \ = 2mathop {lim }limits_{x o infty } frac{{cos frac{{ln frac{{(1 + frac{3}{x})}}{{(1 + frac{1}{x})}}}}{2}frac{1}{2} imes frac{{1 + frac{1}{x}}}{{1 + frac{3}{x}}} imes frac{{ - frac{3}{{{x^2}}}(1 + frac{1}{x}) + frac{1}{{{x^2}}}(1 + frac{3}{x})}}{{{{(1 + frac{1}{x})}^2}}}}}{{ - frac{1}{{{x^2}}}}} hfill \ = 2 hfill \ end{gathered} ]$

這就是考研數學中標準的極限題呀複合函數廣義泰勒展開直接能看出來是2

也可以嘗試用小量寫出來