Question #7b6b6

1 Answer
Apr 16, 2017

#0.#

Explanation:

#"The Limit="lim_(x to 1){(x+2)/(x^2-5x+4)+(x-4)/(3x^2-9x+6)},#

#=lim_(xto1){(x+2)/((x-4)(x-1))+(x-4)/(3(x^2-3x+2))},#

#=lim_(x to 1){(x+2)/((x-4)(x-1))+(x-4)/(3(x-2)(x-1))},#

#=lim_(x to1) {3(x+2)(x-2)+(x-4)^2}/{3(x-4)(x-2)(x-1)},#

#=lim_(x to 1) {3(x^2-4)+(x^2-8x+16)}/{3(x-4)(x-2)(x-1)},#

#=lim_(x to 1) (4x^2-8x+4)/{3(x-4)(x-2)(x-1)},#

#=lim_(x to 1) {4(x^2-2x+1)}/{3(x-4)(x-2)(x-1)},#

#=lim_(x to 1) {4(x-1)^2}/{3(x-4)(x-2)(x-1)},#

#=lim_(x to 1) {4(x-1)}/{3(x-4)(x-2)},#

#={4(1-1)}/{3(1-4)(1-2)},#

#:." The Limit="0.#

Enjoy Maths.!