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Evaluate lim\(_{x→1}\) \(\frac{1-x}{x^2 - 3x + 2}\)

Further Mathematics
WAEC 2025

Evaluate lim\(_{x→1}\) \(\frac{1-x}{x^2 - 3x + 2}\)

  • A. 0
  • B. 1
  • C. -1
  • D. \(\frac{1}{2}\)
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Correct Answer: Option B
Explanation

Given: lim\(_{x→1}\) \(\frac{1-x}{x^2 - 3x + 2}\)

x\(^2\) - 3x + 2

x\(^2\) - 3x + 2 = x\(^2\) - 2x - x + 2

x\(^2\) - 2x - x + 2 = x(x - 2) - 1(x - 2)

= (x - 1)(x -2)

 lim\(_{x→1}\) \(\frac{1-x}{x^2 - 3x + 2}\) = \(\frac{1-x}{(x - 1)(x -2)}\)

= \(\frac{-1}{x -2}\) 

x → 1

\(\frac{-1}{x -2}\) =  \(\frac{-1}{1 -2}\)  =  \(\frac{-1}{-1}\) = 1


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