The concept and properties of higher limit It's too long to remember. What do you mean by a (a ε) and E,

The concept and properties of higher limit It's too long to remember. What do you mean by a (a ε) and E,

Inverted a means: for any given For any given ε, the left e denotes the existence of
This is a common way to define limits

The limit of (1-cos4x) / xsinx when x tends to zero

x -> 0,1- cos4x = 2 sin²2x 8 x² x sinx x²
The original formula is 8

lim xsinx/（arctan3x^2）x→0

x->0,arctanx->x,sinx->x
lim xsinx/（arctan3x^2）
x→0
=lim x²/3x²
x->0
=1/3

lim(1-e^x-x)/(2sinx+xsinx)=0.5lim(1-e^x-x)/x
How does x tend to zero

lim(x->0)[(1-e^x-x)/(2sinx+xsinx)]
=lim(x->0)[(1-e^x-x)/((2+x)sinx)]
=lim(x->0)[(1/(2+x))*(x/sinx)*((1-e^x-x)/x)]
=[lim(x->0)(1/(2+x))]*[lim(x->0)(x/sinx)]*[lim(x->0)((1-e^x-x)/x)]
=[1 / (2 + 0)] * 1 * [LIM (x - > 0) ((1-e ^ x-x) / x)] (apply the important limit LIM (Z - > 0) (Sinz / z) = 1)
=(1/2)*[lim(x->0)((1-e^x-x)/x)]
=0.5*[lim(x->0)((1-e^x-x)/x)].

LIM (e ^ - xsinx) x tends to + infinity

LIM (e ^ - xsinx) x tends to + infinity = LIM (SiNx / e ^ x) x tends to + infinity
Since SiNx is a bounded function, when x tends to + infinity, e ^ x also tends to + infinity
Bounded quantity / Infinity = 0 (take the limit)

LIM (x → 0) [(1 / x ^ 2) - (1 / xsinx)],

1 / x ^ 2-1 / xsinx = (sinx-x) / x ^ 2 * SiNx molecular denominator for derivative (cosx-1) / (2xsinx + x ^ 2 * cosx) or 0 / 0 molecular denominator for derivative - SiNx / (2sinx + 2xcosx + 2xcosx-x ^ 2 * SiNx) in derivation - cosx / (2cosx + 4cosx-4xsinx-2xsinx-x ^ 2 * cosx) x → 0 = - 1 / (2 + 4

Lim x sin (1 / 2x) when x tends to infinity

lim x sin(1/2x)
=lim (1/2)sin(1/2x)/(1/2x)
=1/2

LIM (when x tends to positive infinity) [sin √ (x + 1) - sin √ x]

lim [sin√(x+1)-sin√x]
= lim 2cos( (√(x+1)+√x)/2 ) sin( (√(x+1)-√x)/2 )
For LIM sin ((√ (x + 1) - √ x) / 2)
There is Lim sin ((√ (x + 1) - √ x) / 2) = Lim sin (1 / 2 (√ (x + 1) + x)) = 0
So the primitive can be regarded as a bounded function of infinitesimal multiplication
So the limit is zero

How to calculate the limit Lim X - > infinite [(x + 1) ^ 3] / [(x-1) ^ 2] / x

Divide up and down by X & # 179;
=lim(1+1/x)³/(1-1/x)²
=1³/1²
=1

Lim x →∞ [(x-1) / (x + 3)] ^ (x + 1)
lim x→∞ [(x-1)/(x+3)]^(x+1)
I only know the answer is e ^ (- 4)

lim x→∞ [(x-1)/(x+3)]^(x+1)
=lim x→∞ [1-4/(x+3)]^(x+1)
=lim x→∞ [1-4/(x+3)]^{[-(x+3)/4]*(-4)(x+1)/(x+3)}
=e^{lim x→∞ -4(x+1)/(x+3)}
=e^(-4)