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Mathematics Question on Limits

lim⁡x→1[x+2x2−5x+4+x−43(x2−3x+2)]\lim_{x \to {1}} [\frac {x+2} {x^2 -5x +4} + \frac {x-4} {3(x^2-3x+2)}]

A

0

B

16\frac {1} {6}

C

13\frac {1} {3}

D

1

Answer

0

Explanation

Solution

lim⁡x→1[x+2x2−5x+4+x−43(x4−3x+2)]\lim_{x\to1}\left[\frac{x+2}{x^{2} -5x+4}+\frac{x-4}{3\left(x^{4}-3x +2\right)}\right] = lim⁡x→1[x+2(x−1)(x−2)+x−43(x−1)(x−2)]\lim_{x\to1} \left[\frac{x+2}{\left(x-1\right)\left(x-2\right)}+\frac{x-4}{3\left(x-1\right)\left(x-2\right)}\right] = lim⁡x→1[3(x2−4)+(x−4)23(x−1)(x−2)(x−4)]\lim_{x\to1} \left[\frac{3\left(x^{2}-4\right)+\left(x-4\right)^{2}}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}\right] = lim⁡x→1[4x2−8x+43(x−1)(x−2)(x−4)]\lim_{x\to1}\left[\frac{4x^{2 }-8x +4}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}\right] = 43lim⁡x→1(x−1)23(x−1)(x−2)(x−4)\frac{4}{3} \lim_{x\to1} \frac{\left(x-1\right)^{2}}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)} = 43lim⁡x→1x−1(x−2)(x−4) \frac{4}{3} \lim_{x\to1} \frac{x-1}{\left(x-2\right)\left(x-4\right)}