Question

${\lim }_{x\to 1}[\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}]$

• A. 0
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. 1

Answer 1

Answer: (A)

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