4( 1)x→0 lim(e^x+ cos x+2)/[√( 1-x？ )+ ln( 1-x)]

=4/[ 1+0]=4。

(2) x → 0lim [√ (4+x)-2]/sin (3x), which is a 0/0 limit problem and solved by Roberta's law.

x→0 lim[√( 4+x)-2]/sin(3x)=[√( 4+x)-2]′/[sin(3x)]′

=( 1/2)[√( 4+x)-2](4+x)^(- 1/2)/[3cos(3x)]

=( 1/6)[√( 4+x)-2](4+x)^(- 1/2)/cos(3x)

=0

(3)x→0 lim[( 1-3 x)^(2/sin x)]=( 1-0)^0= 1

(4)x→∞ lim√{x+√[x+√(x+√x)] }-√x→∞

5. Let the function f(x) be continuous at x=2, and f(2)=3.

x→2 lim f(x)[ 1/(x-2)-4/(x？ -4)]=3[(x+2-4)/(x？ -4)]

3[(x-2)/(x？ -4)]=3/(x+2)=3/4

1—— 10

1, the correct proposition is (c)

2，(D)( 1，2)

3. Prove with the mean value theorem.

The mean value theorem includes Rolle theorem, Lagrange mean value theorem and Cauchy mean value theorem, all of which are abstract.

Baidu Wen Ku

Chapter 3: Differential mean value theorem and the application of derivative.

/view/bb 05 ded 85022 AEA 998 f 0 FBC . html