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Question 136 of 148

Q.Evaluate : lim⁡x→1x2−3x+2x2−4x+3\displaystyle\lim_{x\to1}\dfrac{x^2-3x+2}{x^2-4x+3}

Puducherry TnboardTamil Nadu HSC (DGE) Board 2024Subjective· 2mImportance★★★★★
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Cancels the common factor causing the 0/00/0 indeterminate form before substituting the limit point.

  1. Direct substitution of x=1x=1 gives 1−3+21−4+3=00\dfrac{1-3+2}{1-4+3}=\dfrac00, an indeterminate form — factor first.
  2. x2−3x+2=(x−1)(x−2)x^2-3x+2=(x-1)(x-2) and x2−4x+3=(x−1)(x−3)x^2-4x+3=(x-1)(x-3). …

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