Lim(x; 4) (2x^2-7x-4)/(x^3-64) lim(x; -2) (3-sqrt(x+11))/x^2-4 lim(x; 0) (1-cos6x)/x^2

andrewmikilov andrewmikilov    3   10.03.2019 07:00    1

Ответы
sashapeterov85 sashapeterov85  24.05.2020 14:53

im(x;4) (2x^2-7x-4)/(x^3-64)=lim(x;4)(2(x-4)(x+0,5)/((x-4)(x^2+4x+16))=

                                            =2(x+0,5)/(x^2+4x+16)=2(4+0,5)/(4^2+4*4+16)=

                                            =2*4,5/48=9/48=3/16

lim(x;-2) (3-sqrt(x+11))/(x^2-4)=

=lim(x;-2)(3-sqrt{x+11})(3+sqrt{x+11})/((x-2)(x+2)(3+sqrt{x+11}))=

=lim(x;-2)(9-(x+11))/((x-2)(x+2)(3+sqrt{x+11}))=

=lim(x;-2)(-2-x)/((x-2)(x+2)(3+sqrt{x+11}))=

=lim(x;-2)(-(x+2))/((x-2)(x+2)(3+sqrt{x+11}))=lim(x;-2)(-1)/((x-2)(3+sqrt{x+11}))=

=-1/((-2-2)(3+sqrt{-2+11}))=-1/(-4*(3+sqrt{9})=-1/(-4*(3+3))=1/24

 

lim(x;0) (1-cos6x)/x^2-не знаю

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