Question

求解一道高数极限题，在线等

当x趋近于0时，lim（x+sin(x/7))/(x+sin(x/3))
最好能有解答过程，多谢！

Answer 1

求极限x➔0lim[x+sin(x/7)]/[x+sin(x/3)]
解：用等价无穷小作替换：x➔0lim[x+sin(x/7)]/[x+sin(x/3)]＝x➔0lim[x+(x/7)]/[x+(x/3)]
=x➔0lim[(8x/7)/(4x/3)]=6/7
用罗必达求结果一样！
x➔0lim[x+sin(x/7)]/[x+sin(x/3)]＝x➔0lim1+(1/7)cos(x/7)]/[1+(1/3)cos(x/3)]=(1+1/7)/(1+1/3)
=(8/7)/(4/3)=6/7

Answer 2

x趋近于0时，lim（x+sin(x/7))/(x+sin(x/3))

用洛必达法则
lim（x+sin(x/7))/(x+sin(x/3))=lim'（x+sin(x/7))/lim'(x+sin(x/3))=1

追问

老师说期中考试之前的题目不允许用罗比达法则

追答

好吧
lim（x+sin(x/7))/(x+sin(x/3))=lim（1+(sinx/7)/x)/(1+(sinx/3))=lim(1+(1/7)[(sinx/7)/x/7])/(1+(1/3)[(sinx/3)/x/3])=lim(1+(1/7))/(1+(1/3))=6/7
用夹逼定理
洛必达好像错了

Answer 3

由罗比达法则，分别求分子分母的导数。
可得答案15/14

追问

老师说期中考试之前的题目不允许用罗比达法则