 # Lim x → 0 e ^ x-e ^-x-2x / x-sinx limit

## Lim x → 0 e ^ x-e ^-x-2x / x-sinx limit

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### Find the limit Lim x-0 2x / SiNx

Do you mean 2x / SiNx when LIM (x → 0)?
According to the law of lhospital, the numerator and denominator of 2x / SiNx are derived respectively, and
When the original formula = LIM (x → 0), 2 / cosx = 2
Remember that X and SiNx are small quantities of the same order

### Find the limit (x approaches ∞) lim{(x ^ 2 + 2x SiNx) / (2x ^ 2 + SiNx)} My result is 0. What's wrong? Or there are main steps to do the problem

Same above and below divided by X ²
=(1+2/x-sinx/x ²)/ (2+sinx/x ²)
X approaches ∞, so SiN x is bounded, and 1 / X ² Is infinitesimal
Bounded times infinitesimal is infinitesimal
And 2 / X is also infinitesimal
So limit = (1 + 0-0) / (2 + 0) = 1 / 2

### Find the function y = 2x cubic + 3x ²- Extremum and monotone interval of 2x + 10

f(x)=2x^3+3x^2-2x+10
f'(x)=6x^2+6x-2
Let f '(x) = 0
6x^2+6x-2=0
x1=(-3+√21)/6
x2=(-3-√21)/6
f’’(x)=12x+6
F '' (x1) > 0 f (x1) = minimum of 9.71788723
f’’(x2)

### Known function f (x) = 1 3x3 + AX2 + BX (a, B ∈ R) gets the extreme value when x = - 1 (1) Try to express B with an algebraic formula containing a; (2) Find the monotone interval of F (x)

(1) According to the meaning of the question, f '(x) = x2 + 2aX + B. since x = - 1 is an extreme point of the function, f' - 1) = 1-2a + B = 0, B = 2a-1; (2) Because the function f (x) has extreme points, the equation f '(x) = 0 has two unequal real roots. From (1), f' (x) = x2 + 2aX + B = x2 + 2aX + 2A -

### It is known that the function f (x) is a subtractive function on R if f (x ²+ 2)＜f(3x) Find the value range of X

The function f (x) is a subtractive function on R if f (x ²+ 2)＜f(3x)
Then x ² + 2 > 3x
x ² - 3x + 2>0
(x - 1)(x - 2)>0
x> 2 or X