lim (lntan7x)/(ln x趋于0则tan~x且lnx趋于无穷所以原式=limln7x/ln2x=lim(ln7+lnx)/(ln2+lnx)上下除以lnx=lim(ln7/lnx+1)/(ln2/lnx+1)=1上面的的看不懂,怎么直接就等于limlimln7x/ln2x啊,不能理解.=lim (ln sin7x - ln cos7x)/(ln sin2x - ln cos2x)=lim (ln sin7x - 0)/(ln sin2x - 0)=lim (ln sin7x )/(ln sin2x )利用洛笔答法则得=lim (7·cos7x/ sin7x )/(2·cos2x/ sin2x )=lim (7·cos7x·sin2x)/( 2·sin7x·cos2x )=lim (7×1×2x)/( 2×1×7x )=1上面ln tan7x怎么会等于(ln sin7x - ln cos7x)啊,lim(x→0) lntan7x/lntan2x=lim(x→0) (1/tan7x*sec

lim (lntan7x)/(ln
x趋于0则tan~x
且lnx趋于无穷
所以原式=limln7x/ln2x
=lim(ln7+lnx)/(ln2+lnx)
上下除以lnx
=lim(ln7/lnx+1)/(ln2/lnx+1)
=1
上面的的看不懂,怎么直接就等于limlimln7x/ln2x啊,不能理解.
=lim (ln sin7x - ln cos7x)/(ln sin2x - ln cos2x)
=lim (ln sin7x - 0)/(ln sin2x - 0)
=lim (ln sin7x )/(ln sin2x )
利用洛笔答法则得
=lim (7·cos7x/ sin7x )/(2·cos2x/ sin2x )
=lim (7·cos7x·sin2x)/( 2·sin7x·cos2x )
=lim (7×1×2x)/( 2×1×7x )
=1
上面ln tan7x怎么会等于(ln sin7x - ln cos7x)啊,
lim(x→0) lntan7x/lntan2x
=lim(x→0) (1/tan7x*sec²7x*7)/(1/tan2x*sec²2x*2)
=lim(x→0) (7sin4x)/(2sin14x)
=7/2*lim(x→0) (2cos4x)/(7cos14x)
=lim(x→0) cos4x/cos14x
=1/1=1
上面的1/tan7x*sec²7x*7为什么等于7sin4x啊