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When x → 0, Lim [e ^ x + (e ^ - x) - 2] / SiNx ^ 2
lim(x->0) lim[e^x+(e^-x)-2]/(sinx)^2 (0/0)=lim(x->0) lim[e^x-(e^-x)]/(sin2x) (0/0)=lim(x->0) lim[e^x+(e^-x)]/(2cos2x)= 2/2=1
lim(1-e^(sinX))^(1/x) X approaches 0
Take logarithm to get:
lim ln（1-e^(sinx))/x
Use lobidad once
lim -e^(sinx)*cosx/(1-e^(sinx))
Again, lobidad
lim -e^(sinx)*cosx*cosx+e^(sinx)*sinx /(-e^(sinx)*cosx)
About e ^ (SiNx)
Lim - cosx ^ 2 + SiNx / cosx is the limit
-1
So the original limit is 1 / E
LIM (x tends to 0) (1-cosx ^ 2) / ((x ^ 3) * SiNx)
LIM (x tends to 0) (1-cosx ^ 2) / ((x ^ 3) * SiNx)
=LIM (x tends to 0) [(x ^ 2) ^ 2] / 2 / ((x ^ 3) * x)
=1/2 lim x^4/x^4
=1/2
1. LIM (when x tends to 0) [x ^ 2 (1-cosx)] / [(1 + e ^ x) (SiNx) ^ 3] This limit is equal to 0, but I don't know how this part of (1 + e ^ x) turns into no arrow! Teach me
Not simplified to no, (1 + e ^ x) is 2, 1-cosx is equivalent to x ^ 2 / 2, the molecule has the 4th power of X, and the denominator (SiNx) ^ 3 is equivalent to the 3rd power of X. the molecule is a higher-order infinitesimal, so the limit is 0
The function f (x) = x + A / x, a > 0 is known. If f (1) = f (2), it is proved that f (x) is monotonically decreasing on (0,2] How can I prove that it is monotonous
A = 2 can be obtained from F (1) = f (2), so f (x) = x + 2 / X,
Derivative: F '(x) = 1-2 / x ^ 2 = (x ^ 2-2) / x ^ 2
When x belongs to (0, radical 2], f '(x) < 0, so f (x) is minus,
So this question should be the wrong interval. It should be (0, radical 2)
Given the function f (x) = |x-1| (x + 3), (1) find the monotone interval of function f (x), and prove the monotone decreasing interval; (2) Find the maximum value of function f (x) on interval [- 3,0]
When x ≥ 1, f (x) = (x-1) (x + 3) = (x + 1) ²- 4 it is a subtractive function on (- ∞, - 1], and an increasing function on [- 1, + ∞). Because x ≥ 1, f (x) is an increasing function when x ≥ 1. When x ≤ 1, f (x) = - (x-1) (x + 3) = - (x + 1) ²+ 4 it is an increasing function on (- ∞, - 1], and it is an increasing function on [- 1, + ∞
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