幾道微積分求極限,求大神,文科妹子很憂傷啊

2022-05-26 06:56:33 字數 2340 閱讀 6459

1樓:匿名使用者

第一張紙

(1)lim(x->0) (a^x -b^x) /x

=lim(x->0) [(1+xlna) -(1+xlnb)] /x

=lna -lnb

(2)lim(x->0) [cosx -√(1+x) ]/x

=lim(x->0) /x

=-1/2

(3)x->0

e^x -e^(-x) -2x

~ ( 1+x +(1/2)x^2 + (1/6)x^3) -( 1-x +(1/2)x^2 - (1/6)x^3) -2x

~(1/3)x^3

------

sinx ~ x- (1/6)x^3

x-sinx ~ (1/6)x^3

-------

lim(x->0) [e^x-e^(-x) -2x ]/(x-sinx)

=lim(x->0) (1/3)x^3/[(1/6)x^3]

=2(4)

a^x ~ 1+ x(lna) +(1/2) x^2. (lna)^2 + (1/6)x^3 .(lna)^3

a^(sinx)

~ a^(x - (1/6)x^3)

~ 1+ (x-(1/6)x^3)(lna) +(1/2) (x-(1/6)x^3)^2. (lna)^2 + (1/6)(x - (1/6)x^3)^3 .(lna)^3

~ 1+ x(lna) +(1/2)x^2.(lna)^2 +[(-1/6)lna +(1/6)(lna)^3 ]x^3

a^x - a^(sinx)~ -(1/6)(lna).x^3

----------

lim(x->0) [a^x-a^(sinx)]/(sinx)^3

=lim(x->0) -(1/6)(lna)x^3/x^3

=-(1/6)(lna)

第2張紙

(1)lim(x->+∞) lnx/x^n (∞/∞)

=lim(x->+∞) (1/x)/[nx^(n-1)]

=lim(x->+∞) 1/[nx^n]

=0(2)

lim(x->0+) lnsinx/ln(sin√x) (0/0)

=lim(x->0+) [cosx/sinx]/[ (cos√x/sin√x) .(1/(2√x)) ]

=lim(x->0+) 2√x tan√x / tanx

=lim(x->0+) 2x /x

=2(3)

lim(x->+∞) e^x/lnx (∞/∞)

=lim(x->+∞) e^x/(1/x)

=lim(x->+∞) x.e^x

->∞(4)lim(x->+∞) (e^x-x)/(e^x +x)

=lim(x->+∞) [1 - 2x/(e^x +x)]

=1 - lim(x->+∞) 2x/(e^x +x)]

= 1- 0

=1(5)

lety=1/x

lim(x->+∞) x[e^(1/x) -1]

=lim(y->0) (e^y -1)/y

=lim(y->0) y/y

=1(6)

lety=x-1

y->0

(1+y)ln(y+1) ~ y+y^2

(1+y)ln(y+1) -y ~ y^2

------

lim(x->1) [ x/(x-1) - 1/lnx )

=lim(y->0) [ (1+y)/y - 1/ln(y+1) ]

=lim(y->0) [ (1+y)ln(y+1) -y ] /[y.ln(y+1) ]

=lim(y->0) y^2/y^2

=1(7)

lety= x-π/2

lim(x->π/2) ( secx -tanx)

=lim(x->π/2) ( 1/cosx -tanx)

=lim(y->0) ( 1/siny -coty)

=lim(y->0) ( 1-cosy)/siny

=lim(y->0) ( 1/2)y^2)/y

=0(8)

l =lim(x->0+) (sinx)^[2/(1+lnx)]

lnl=lim(x->0+) 2ln(sinx)/(1+lnx) (∞/∞)

=lim(x->0+) 2cotx/(1/x)

=2lim(x->0+) x/tanx

=2lim(x->0+) 2ln(sinx)/(1+lnx) =l = e^2

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