Question

高数用洛必达求极限

高数用洛必达求极限如图这两题 请给出具体步骤

Answer 1

如图

Answer 2

(1)
lim(x->0) ln(1+x^2)/(secx -cosx)
=lim(x->0) cosx. ln(1+x^2)/[1 -(cosx)^2 ]
=lim(x->0) ln(1+x^2)/[1 -(cosx)^2 ]
=lim(x->0) ln(1+x^2)/ (sinx)^2
=lim(x->0) x^2/ x^2
=1
(2)
lim(x->0) x^2 . e^(1/x^2)
=lim(x->0) e^(1/x^2) /(1/x^2 ) (∞/∞)
=lim(x->0) (-2/x^3).e^(1/x^2) /(-2/x^3 )
=lim(x->0) e^(1/x^2)
->∞

Answer 3

等价无穷小ln(1+u)~u，sinu~u
=limx²cosx/(1-cos²x)=1
u=1/x²趋于+∞
=lime^u/u=lime^u/1=+∞