lim(x→0+)[ln(sin5x)]/[ln(sin3x)] 这题咋做

lim(x→0+)[ln(sin5x)]/[ln(sin3x)]
=lim(x→0+)[5cos5x/sin5x]/[3cos3x/sin3x]
=lim(x→0+)[5cos5x×sin3x]/[3cos3x×sin5x]
=1（L'Hospital法则）

补充：lim(x→0+)cos5x=1,lim(x→0+)cos3x=1，因为这是连续函数求极限,直接代入。

lim(x→0+)[sin3x/sin5x]
=3/5

lim(x→0+)[ln(sin5x)]/[ln(sin3x)]
=lim(x→0+)[5cos5x/sin5x]/[3cos3x/sin3x]
=lim(x→0+)[5cos5x/3cos3x]*[sin3x/sin5x]
=lim(x→0+)[5cos5x/3cos3x]*(x→0+)[sin3x/sin5x]
=5/3*3/5
=1

sin5x~5x sin3x~3x
因此原式=lim(x→0+)ln(5x)/ln(3x)=lim(ln5+lnx)/(ln3+lnx)
=lim(ln5/lnx+1)/ln3/1nx+1)
因为x→0+ 故lnx→-∞
那么ln5/lnx,ln3/lnx都趋近于0
故
原式=1