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
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
由罗比达法则，分别求分子分母的导数。
可得答案15/14
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