What is the limit of SiNx over x when x approaches 0?

What is the limit of SiNx over x when x approaches 0?

one

When x approaches 0, the limit value of x ^ SiNx is obtained

Limx - > 0 x ^ SiNx = limx - > 0 e ^ LNX ^ SiNx = limx - > 0 e ^ (sinxlnx) = limx - > 0 e ^ (LNX / 1 / SiNx) index is of ∞ / ∞ type

Find the limit of (a ^ x-a ^ SiNx) / (x ^ 2 * SiNx) when x approaches 0
Wrong denominator! The square should be SiNx, not x ^ 2

The answer is 1 / 6 * ln (a), which can be solved by Taylor series
a^x=e^(ln(a)*x)=ln(a)*x+1/2*(ln(a)*x)^2+1/6*(ln(a)*x)^3+o(x^3)
a^sin(x)=e^(ln(a)*sin(x))=ln(a)*sin(x)+1/2*(ln(a)*sin(x))^2+1/6*(ln(a)*sin(x))^3+o(x^3)
(substitute sin (x) = X-1 / 6 * x ^ 3 + O (x ^ 3), omit more than three times)
=ln(a)*x-1/6*ln(a)*x^3+1/2*ln(a)^2*x^2+1/6*ln(a)^3*x^3+o(x^3)
So subtracting a ^ x-a ^ sin (x) = 1 / 6 * ln (a) * x ^ 3 + O (x ^ 3),
Denominator = x ^ 3 + O (x ^ 3),
So the answer is 1 / 6 * ln (a)

The limit of minus one square of 2 when x tends to zero
It's 2 ^ (- 1 / x ^ 2)

Solution:
lim[x->0]2^(-1/x^2)
=2^lim[x->0] -1/x^2
=2^(-1/(lim[x->0]x^2))
=2 ^ (- 1 / positive infinity)
=2^0
=1

Find the limit of subtracting the square of the cotangent of X from the square of X when x approaches zero

Limit[1/x^2-(Tan[(Pi/2)-x])^2,x->0]=2/3

What's the square of the integral sign. SiNx? DX =,

The square of SiNx is (SiNx) ^ 2
(sinx)^2=(1-cos2x)/2
∫sinx)^2dx
=∫(1-cos2x)/2dx
=1/2*[∫dx-∫cos2xdx]
=1/2*[∫dx-1/2*∫cos2xd2x]
=x/2-1/4*sin2x+c

The detailed steps of the integration of the square of X and SiNx

Let J be the sign of integral and d be the sign of differential, that is, J (x ^ 2 * SiNx) DX = J (x ^ 2) d (- cosx) = - x ^ 2 * cosx-j (- 2x * cosx) DX = - x ^ 2 * cosx + J (2x) d (SiNx) = - x ^ 2 * cosx + 2x * SiNx - J (2sinx) DX = - x ^ 2 * cosx + 2x * SiNx + 2cosx + C

Why is the limit of SiNx 1 when x tends to zero

When x tends to zero, the limit of SiNx is 0

What is the limit of the absolute value of SiNx when x tends to zero negative?

X tends to zero, which means x tends to zero from the left
When x tends to 0 from the left, Lim | SiNx | = - limsinx = - 0 = 0

When x approaches zero, the limit of SiNx / 6x

Solution
lim 1/6 sinx/x(x->0)=1/6
It's based on two important limits
lim(x_ >0)sinx/x=1