Using the law of l'urbida to find the limit, LIM (x tends to 0) [the x power of E - the negative x power of e] / X

Using the law of l'urbida to find the limit, LIM (x tends to 0) [the x power of E - the negative x power of e] / X

LIM (x tends to 0) [e's x power - E's negative x power] / x = LIM (x tends to 0) [e's x-th power) / x-lim (x tends to 0) [e's negative x-th power) / x = LIM (x tends to 0) [e's X-Power) + LIM (x tends to 0) [e's negative X-Power] / x = LIM (x tends to 0) [e's negative X-Power] / x = 1 + 1 = 2

How to calculate the SiNx power of X when the limit LIM (x tends to 0 +) is calculated by the high number lobida's rule?

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What is the limit LIM (x approaches to 0) (the third power of e-The negative x-4x of E) / 1-cosx?

lim(x->0) (e^3-e^(-x) -4x)/(1-cosx)
= lim(x->0) [ e^(-x) - 4) / sinx
= (1- 4) /1
= -3

Find the limit LIM (SiNx / x) ^ (1 / 1-cosx), X tends to 0 The answer is minus one third of E Why can't we use the equivalent substitution formula to get SiNx / x = 1? Then the limit is equal to 1?

Because SiN x / X only takes the limit value when x tends to 0, the limit of exponent is ∞. The ∞ power of limit 1 is infinitive. LIM (SiNx / x) ^ (1 / 1-cosx) = e ^ LIM (1 / 1-cosx) · ln (SiNx / x) = e ^ Lim (1 / (x? 2)) · ln (1 + SiNx / X - 1) [equivalent substitution: 1-cosx ~ x
... "

Find the limit LIM (cosx + SiNx) ^ 1 / X X tends to 0 It means that I haven't learned the law of rabida

Let a = LIM (cosx + SiNx) ^ 1 / x, then LNA = Lim ln (cosx + SiNx) / x = Lim [ln (cosx + SiNx)] '/ X' [L'Hospital Rule] = LIM (cosx + SiNx) '/ (cosx + SiNx) = LIM (- SiNx + cosx) / (cosx + SiNx) = LIM (0 + 1) / (1 + 0) = 1

Find the limit of LIM (1 + cosx) ^ (2secx), X → π / 2

Original formula = e ^ lim2secxln (1 + cosx)
=e^lim2secxcosx
=e^2
When x tends to π / 2, cosx tends to 0, so ln (1 + cosx) is equivalent to cosx
Therefore, it is good to replace ln (1 + cosx) with cosx in limit operation

Urgent! Find the limit of LIM (1-cosx) ^ (2secx), X → π

Since 1-cos π and 2sec π are not equal to 0, they are not indefinite and can be directly substituted into π calculation
The original formula = (1-cos π) ^ (2sec π)
=[1-(-1)]^[2(-1)]
=2^(-2)
=1/4

Find the value range of x power + 1 of X power-3 of function y = 9

Let t = 3 to the power of X, t ∈ (0, + ∞)
y=t^2-t+1=(t-1/2)^2+3/4
T ∈ (3 / 4, + ∞) is obtained by the combination of number and shape

The value range of y = 1 / (x power of 3 + 1) is

X power of 3 + 1 > 0 + 1 = 1
therefore
y＜1/1=1
And Y > 0
therefore
The range is (0,1)

Let x + 1 power + 3, X belong to the range of (negative infinity, 1]

Let a = 2 ^ X
Then 4 ^ x = a
2^(x+1)=2a
X