Question

求

lim

x→0

ex-e-x-2x

x-sinnx

的值．

Answer 1

分析：由羅比塔法則求解．解題時(shí)要注意當(dāng)n≠1時(shí)和當(dāng)n=1時(shí)的不同結(jié)果．

解答：應(yīng)用羅比塔法則，

lim

x→0

ex-e-x-2x

x-sinnx

=

lim

x→0

ex-e-x-2

1-ncosnx

=0．(n≠1)
當(dāng)n=1時(shí)，

lim

x→0

ex-e-x-2x

x-sinnx

=

lim

x→0

ex-e-x-2

1-cosx

=

lim

x→0

ex-e-x

sinx

=

lim

x→0

ex-e-x

cosx

=2

點(diǎn)評(píng)：本題考查極限的性質(zhì)和應(yīng)運(yùn)，解題時(shí)要注意公式的靈活運(yùn)用．