Calculus Let f (0) = 1. G (1) = 2, f '(0) = - 1, G' (1) = - 2 find: 1、limx→0[cosx-f(x)]/x； 2、limx→0[2^x*f(x)]/x； 3. Limx → 1 [root x * g (x) - 2] / X-1 Correction of question 2 2、limx→0[2^x*f(x)-1]/x

Calculus Let f (0) = 1. G (1) = 2, f '(0) = - 1, G' (1) = - 2 find: 1、limx→0[cosx-f(x)]/x； 2、limx→0[2^x*f(x)]/x； 3. Limx → 1 [root x * g (x) - 2] / X-1 Correction of question 2 2、limx→0[2^x*f(x)-1]/x

1. Take the upper and lower derivative - sinx-f '(x) and bring x = 0 in. The result is 1
I don't know if you made a mistake. Just bring x = 0 in. The numerator is 1 and the denominator is 0. The result is positive infinity
3. I have to think about it
2. The result of upper and lower derivation is 2 ^ xln2f '(x). Just substitute x = 0 and the result is - LN2

On mathematical problems of calculus ∫（arctanx）^2/1+X^2 dx

∫（arctanx)^2/(1+x^2) dx
=∫（arctanx)^2 d(arctanx)
=(arctanx)^3/3 + C

Mathematical problems of calculus F '(x) = 1 / x ^ 2 find f (x)? I forgot all about it

-1/x+c

About calculus The radius r of water ripple increases at the rate of 0.35m per second. Calculate the area change speed of the circle when the radius is r = 1m Why DS / dt = 2 * 3.14 * r * DR / dt ds/dr=2*pi*r dr/dt=0.35 ds/dt=ds/dr * dr/dt =2*pi*r*0.35=6.28*0.35

You're not a college student, are you?
Just find the derivative
s=3.14r^2
ds/dt=6.28*r*dr/dt
=6.28*1*0.35
=2.198(m^2/s)

The problem of finding limit by calculus When x 0, f (x) = (1-cosx) in (1 + 2x) and (?) are infinitesimals of the same order A x3 B x4 Cx5 Dx6

A. 1-cosx is x ^ 2 / 2 and ln (1 + 2x) is 2x, so choose a

The problem of limit in Calculus Lim x 3 arcsin (x ^ 2-9) / (x-3)

This is an ∞ / ∞ infinitive,
lim (x→3) arcsin(x^2-9)/(x-3）
=lim (x→3)〖arcsin(x^2-9)〗’/〖(x-3)〗’
=lim (x→3)2x/∫1-(x^2-9)^2
=6 (bring 3 in)