A problem in advanced mathematics should be processual and standardized.

Question 1: The tangent at the point (0, f(x)) is also the tangent at the point (0, f(0)).

x 0, lim f(x) 3x =1 , then f(x) and 3x are equivalent infinitesimal and f(x) is continuous at point 0 then there must be f(0)=0, then it is the tangent at the point (0,0), now only the slope k=f is required'(0) is sufficient.

f'(0)= lim [f(0+△x) -f(0)]/△x = lim f(△x)/△x --x→0.

Substituting the equivalent infinitesimal 3 x for f(x) yields:

f'(0)= lim 3△x /△x = 3

Then the tangent is: y = 3x

Question 2: Since you are asking about the conditions, then you must emphasize the sufficiency and necessity.

What you say f(x)=0 satisfies sufficiency, but not necessity.

It is perfectly possible to find the : directly from the definition of the derivative

-x→0f'(0) = lim [ f(0+△x) -f(0)]/△x

lim [（1+tan|0+△x|）*f(0+△x） -1+tan|0|)f(0)]/△x

lim [ 1+tan|△x|) f(△x) -f(0)]/△x

To make this exist then: f(0)=0

Conversely, if f(0)=0, f(x) is a derivative function then.

f'(0) lim [ 1+tan|△x|) f(△x) -f(0)]/△x

lim [(1+tan|△x|) f(△x) ]/△x

f'(0) *lim(1+tan|△x|）

f'(0) So instead of f(x)=0, just make f(0)=0.

1) f(x) is continuous at x=0 and tends to 0 at x, so when x tends to 0, f(x) tends to infinitesimal, when x tends to 0, lim f(x) 3x =1, the left denominator numerator is infinitesimite, and when x tends to 0, limf can be obtained when x tends to 0'(x) 3=1, f'(x) The derivative value at x 0 is 3

So the slope of the tangent is k 3, and after the (0,0) point, the tangent equation is y 3x

2) f(x) must be derivable at 0 at x=0+, x=0- on both sides, and must be defined in this case.

The tangent at the point (0,f(x)) is also the tangent at the point (0,f(0)).

x 0, lim f(x) 3x =1 , then f(x) and 3x are equivalent infinitesimal and f(x) is continuous at point 0 then there must be f(0)=0, then it is the tangent at the point (0,0), now only the slope k=f is required'(0) can be f'(0)= lim [f(0+△x) -f(0)]/△x = lim f(△x)/△x --x→0.

Substituting the equivalent infinitesimal 3 x for f(x) yields:

f'(0) = lim 3 x x = 3 then the tangent is: y = 3x

The second question can be done directly with definitions.

The integral field d is the circle (x-1) 2+y 2 = 4 i.e. the upper left quarter circle of r = 2cost.

f(x,y)dxdy

<0,π/4>dt∫<0, 1/cost>f(rcost,rsint) rdr

dt∫<0, 2cost>f(rcost,rsint) rdr

Select c to rotate around the y-axis, y does not move, x2 becomes x2+z2

Hello! Only A is correct.

a1.Arbitrarily adding or removing the finite nature of the series does not change its convergence.

2.If the series an converges and the series bn converges, then the series (an+bn) also converges.

The general term is split into two parts, un and u(n+1), it is known that un converges, and u(n+1) is only one less u1 than un, and removing the finite term of the series does not change the convergence, so u(n+1) also converges, and then uses the properties of the series, (un+u(n+1)) converges.

b counterexample (1) un=(-1) n n

un converges, and u(2n) diverges.

c counterexample un=(-1) n n

un*u(n+1)=(-1) n n*(-1) (n+1) (n+1)=-1 ( n (n+1)) divergence.

d counterexample un=(-1) n n,(-1) n*(-1) n n=1 n divergence.

If there is anything you don't understand, you can ask at any time, I will try my best to answer, I wish you academic progress, thank you. xd

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