Tính giới hạn: $lim{x->0}\frac{lnsin4x}{lnsin5x}$ 04/07/2021 Bởi Ruby Tính giới hạn: $lim{x->0}\frac{lnsin4x}{lnsin5x}$
Lời giải: Ta có: $lim{x->0}\frac{lnsin4x}{lnsin5x}$$=lim{x->0}\frac{4cotg4x}{5cotg5x}$$=lim{x->0}\frac{4}{5}.\frac{tg5x}{tg4x}$$=lim{x->0}\frac{4}{5}.\frac{5x}{4x}$$=lim{x->0}\frac{4}{5}.\frac{5}{4}$$=1$ Bình luận
`lim_{x -> 0} (lnsin 4x)/(ln.sin 5x)` `= lim_{x -> 0} (4.cot 4x)/(5.cot 5x)` `(text{Quy tắc L’Hospital})` `= lim_{x -> 0} (4.tan 5x)/(5.tan 4x)` `= 4/(5).(5)/4` `= 1` Bình luận
Lời giải:
Ta có:
$lim{x->0}\frac{lnsin4x}{lnsin5x}$
$=lim{x->0}\frac{4cotg4x}{5cotg5x}$
$=lim{x->0}\frac{4}{5}.\frac{tg5x}{tg4x}$
$=lim{x->0}\frac{4}{5}.\frac{5x}{4x}$
$=lim{x->0}\frac{4}{5}.\frac{5}{4}$
$=1$
`lim_{x -> 0} (lnsin 4x)/(ln.sin 5x)`
`= lim_{x -> 0} (4.cot 4x)/(5.cot 5x)` `(text{Quy tắc L’Hospital})`
`= lim_{x -> 0} (4.tan 5x)/(5.tan 4x)`
`= 4/(5).(5)/4`
`= 1`