Help with these questions about the limits of mathematics

1)(x->0)lim(1+2x)^x\1

x->0)lim（1+2x)^（2x\2）

x->0)(lim(1+2x)^ 2x\1))^2

x->0)lim(1+x)^(x\1)=e

x->0)(lim(1+2x)^ 2x\1))^2=e^2

2)(x->+limx(ln(x+2)-lnx)

x->+limxln((x+2)/x)

x->+limln(1+2/x)^x

x->+limln(1+1/(x/2))^x/2)2

x->+2ln(lim(1+1/(x/2))^x/2))2lne

3) The question should be x->0

x->0)lim(((1+x+x^2)^1/2-1)/sinx)

x->0)lim(1 2(1+x+x 2) (1 2) cosx) (according to Lobida's rule, simultaneous derivative).

4) The numerator and denominator both tend to infinity, and according to Lobida's law, the derivatives are obtained separately.

x->+lim(2ax+b)=5

a=0,b=5

5)(x->2)lim(ax+b)/(x-2)

x->2)lim(a(x-2)/(x-2)+(2a+b)/(x-2))

x->2)lim(a+(2a+b)/(x-2))

a+(x->2)lim((2a+b)/(x-2))

a=3,b=-6

6)f(x)=(x^2-4)^(1/2)

x^2-4>0

x^2>4

x>2∪x<-2

Continuous interval (- 2) (2, ).

7) Let f(x) be defined in a hollow neighborhood of point x f(x) is discontinuous at point x, that is, "f(x)=f(x)" is not true, and the following three situations may occur:

1) f(x) is not defined in x=x;

2) f(x) is defined in x=x, but f(x) does not exist;

3) f(x) is defined in x=x, and f(x) also exists, but f(x) ≠ f(x).

X in one of the three cases is a discontinuous point of f(x) and is called a discontinuity point

For example: f(x)=(x 2-1) x(x-1)=(x+1)(x-1) x(x-1).

x(x-1)=0

x=0,x=1

x=0 and x=1 are the discontinuities of the function.

f(x) is not defined at x=0.

x=0 belongs to the first case.

x=1, f(x)=0, f(x) is defined at x=1, and f(x) exists.

But (x->1)limf(x)=2≠f(1).

x=1 belongs to the third case, which is the discontinuity point.

1. The numerator and denominator are physical and chemical.

lim(x+x^2)(√1+x)+√1-x))/2x(√(1+x+x^2)+1)=1/2

2、lime^xlnx

limxlnx=limlnx/(1/x)=lim(1/x)/(1/x^2)=lim(-x)=0

limx^x=e^xlnx=1

3. Let t=1 x

lim[t-ln(1+t)]/t^2 t-->0=lim[1-1/(1+t)]/2t

Is there an obvious mistake in the first one, whether it is positive infinity or negative infinity, it is meaningless.

The second answer should be 1

What is the third base, please ask you this kind of question.

This topic should be "sum difference product".

sinx-siny=2cos((x+y)/2)*sin((x-y)/2)

So sinx-sina = 2cos((x+a) 2)*sin((x-a) 2).

So) (sinx-sina) (x-a) =2cos((x+a) 2)*sin((x-a) 2) (x-a) ;

It is easy to know that lim(x->0)(sinx x)=1;

So lim(x->a)(sinx-sina) (x-a).

lim(x->a)/2

2*cosa/2=cosa;

You can also use the geometric method!

lim(x-sina) (x-a) represents the tangent of the x=a point on the image of the function y=sinx, so.

cosa;

Find its original function.

i.e. the inverse process of derivation.

With the law of Lobida, if you don't know how to do it, there is an encyclopedia.

Will the sum difference product of trigonometric functions?

I don't write LIM, and at a certain point, I go straight in.

1) Original formula = (x+2) (x-1) (x(x-1)) = (x+2) x=3

2) Original formula = (x+2) (x 2-2x+4) ((x+2)(x-6)) = (x 2-2x+4) (x-6) = (4+4+4) (-8) = -3 2

3) Original formula = (4-3 x+2 x 2) (-1+1 x+1 x 2) = (4+0+0) (-1+0+0) = -4

4) Original formula = (x+4-3 2) ((x-5)( x+4)+3))=1 ((x+4)+3)=1 6

1) It can only be said that the generalized limit is +, because 1 x 2>0 and can be very ......

2) Original formula = (x+1+1 x) (1-1 x)=x=-

1) Original formula = sin(3x) (3x)*3=1*3=3

2) Original formula = sin(3x) (3x)*(5x) sin(5x)*3 5=1*1*3 5=3 5

3) Original formula = xsinx (2sin 2(x 2)) = x 2 (2(x 2) 2) = 2

The last 3 questions are:

Question 1 3 Question 2 3 5

Big brother: It's not that you won't do it, it's because you're lazy.

All are of the "0 0" type, and the Robida rule can be used:

Ahh