Вычислите пожалуйста придел функции:
x² - 25
1) lim -------------
x⇒-5 x+5
2x+x²
2) lim ------------
x⇒-2 x²+5x+6
x²-3x
3) lim ------------
x⇒∞ x²-8
im[x->0] sin(x) = x - x^3/6 + lim[x->0] cos(x) = 1-x^2/2 + lim[x->0] tg(x) = x+ 1/3*x^3 + lim[x->0] ( tg(x)-sin(x) ) = (1/3+1/6)*x^3 = 1/2*x^3 lim[x->0] (1-cos(a*x) ) = 1/2*a^2*x^2 lim[x->0] ( tg(x)-sin(x) ) /x/(1-cos(a*x) ) = 1/a^2 2. tg(x)-sin(x) = sin(x)/cos(x)*(1-cos(x)) = sin(x)* 2*sin^2(x/2)/cos(x) 1-cos(a*x) = 2*sin^2(a/2*x)^2 lim[x->0] sin(x)/x ( sin(x/2)/x/2)^2 * ( a*x/2/sin(a*x/2))^2 1/cos(x)*1/a^2 = 1/a^2
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
im[x->0] sin(x) = x - x^3/6 +
lim[x->0] cos(x) = 1-x^2/2 +
lim[x->0] tg(x) = x+ 1/3*x^3 +
lim[x->0] ( tg(x)-sin(x) ) = (1/3+1/6)*x^3 = 1/2*x^3
lim[x->0] (1-cos(a*x) ) = 1/2*a^2*x^2
lim[x->0] ( tg(x)-sin(x) ) /x/(1-cos(a*x) ) = 1/a^2
2.
tg(x)-sin(x) = sin(x)/cos(x)*(1-cos(x)) = sin(x)* 2*sin^2(x/2)/cos(x)
1-cos(a*x) = 2*sin^2(a/2*x)^2
lim[x->0] sin(x)/x ( sin(x/2)/x/2)^2 * ( a*x/2/sin(a*x/2))^2 1/cos(x)*1/a^2 = 1/a^2