用洛必达法则求下列极限,急用,

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(lntan7x)'
=7[1/(tan7x)](sec²7x)
(lntan2x)'
=2[1/(tan2x)](sec²2x)
因此：原极限=(7/2)lim(x→0) tan2x/tan7x
=(7/2)lim(x→0) (2sec²2x)/(7sec²7x)
=lim(x→0) sec²2x/sec²7x
=1