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Q. $\lim_{x \to {1}} [\frac {x+2} {x^2 -5x +4} + \frac {x-4} {3(x^2-3x+2)}]$

Solution:

$\lim_{x\to1}\left[\frac{x+2}{x^{2} -5x+4}+\frac{x-4}{3\left(x^{4}-3x +2\right)}\right]$
= $\lim_{x\to1} \left[\frac{x+2}{\left(x-1\right)\left(x-2\right)}+\frac{x-4}{3\left(x-1\right)\left(x-2\right)}\right]$
= $\lim_{x\to1} \left[\frac{3\left(x^{2}-4\right)+\left(x-4\right)^{2}}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}\right]$
= $\lim_{x\to1}\left[\frac{4x^{2 }-8x +4}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}\right]$
= $\frac{4}{3} \lim_{x\to1} \frac{\left(x-1\right)^{2}}{3\left(x-1\right)\left(x-2\right)\left(x-4\right)}$
= $ \frac{4}{3} \lim_{x\to1} \frac{x-1}{\left(x-2\right)\left(x-4\right)}$