пожалуйста помогите решить lim x->4 (3-√(5+x)) / (1-√(5-x)) срочноооо. Created by mili96. matematika-ru

(1-√(5-x)) срочноооо...

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пожалуйста помогите решить

lim x->4 (3-√(5+x)) / (1-√(5-x))

срочноооо

Answers & Comments

zoloto100 1)
lim x -> 1 (x^3 - 1)/(5x^2 - 4x - 1) = lim x -> 1 (x - 1)(x^2+x+1)/5(x-1)(x+1/5) = lim x -> 1 (x^2+x+1)/5(x+1/5) = 3/(5*1.5) = 2/5
2)
lim x -> oo (5 - 2x - 3x^2)/(x^2+x+3) =lim x -> oo (5/x^2 - 2/x - 3)/(1 + 1/x + 3/x^2) = -3/1 = -3
3)
lim x -> oo (2x - sqrt(4x^2 + 3x)) = {Дамножим и поделим на сопряженное значение}
lim x -> oo (4x^2 - (4x^2 + 3x)) / (2x + sqrt(4x^2 + 3x)) =
lim x -> oo -3x/(2x + sqrt(4x^2 + 3x)) =
lim x -> oo -3/(2 + sqrt(4 + 3/x))= -3/4
4)
lim x -> 0 (cos x - 1)/x^2 =
=lim x -> 0 2*sin^2(x/2)/ x^2 =
=lim x -> 0 2 sin^2(x/2)/(4* (x/2)^2) =
lim x -> 0 0.5 (sin x/2 / x/2)^2 =
{в скобках первый замечательный предел}=
0.5*1= 0.5
5)
lim x-> 0 = (sin x - tg x)/x = lim x->0 sinx /x - (sinx/cosx) 1/x =
lim x->0 sinx /x - (sinx /x) 1/cosx =
lim x->0 sinx /x - lim x->0 (sinx /x) lim x->0 1/cosx =
1 - 1 = 0
6)
lim x -> oo (1 + 2/3x)^x =
= lim x-> oo ((1 + 1/(3x/2))^(3x/2))^2/3 =
{Второй замечательный предел} =
e ^ 2/3
7)
lim x -> oo [2x/(2x+1)]^x =
lim x -> oo (1/ (2x+1)/2x) ^x =
lim x -> oo (1/ (1 + 1/(2x)))^x =
lim x-> oo 1/ (1 + 1/2x)^x =
lim x-> oo 1/ ((1 + 1/2x)^2x)^0.5 =
1/e^0.5

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mili96 а когда x->4

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