求极限:lim(x→0+)ln(sinax)/ln(sinbx) (a>0,b>0)
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求极限:lim(x→0+)ln(sinax)/ln(sinbx) (a>0,b>0)
![求极限:lim(x→0+)ln(sinax)/ln(sinbx) (a>0,b>0)](/uploads/image/z/5021340-60-0.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90%EF%BC%9Alim%28x%E2%86%920%2B%29ln%28sinax%29%2Fln%28sinbx%29+%28a%3E0%2Cb%3E0%29)
不定式,最好用洛必达法则上下分别求导了
lim(x→0+) ln(sinax)/ln(sinbx)
=lim(x→0+) (1/sinax*acosax)/(1/sinbx*bcosbx),用洛必达法则
=(a/b)lim(x→0+) tanbx/tanax
=(a/b)lim(x→0+) [bsec^2(bx)]/[asec^2(ax)],用洛必达法则
=(a/b)(b/a)lim(x→0+) cos^2(ax)/cos^2(bx)
=1*1/1
=1
lim(x→0+) ln(sinax)/ln(sinbx)
=lim(x→0+) (1/sinax*acosax)/(1/sinbx*bcosbx),用洛必达法则
=(a/b)lim(x→0+) tanbx/tanax
=(a/b)lim(x→0+) [bsec^2(bx)]/[asec^2(ax)],用洛必达法则
=(a/b)(b/a)lim(x→0+) cos^2(ax)/cos^2(bx)
=1*1/1
=1
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