Its the limit as x tends to 3

Mathematics

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My apologies for the oversight! Thank you for clarifying the limit value. Let's re-evaluate the limit as xx approaches 3.

ii. Evaluate lim⁡x→3x2−9x2+2x−3\lim_{x \to 3} \frac{x^2 - 9}{x^2 + 2x - 3}

  1. Check the form of the limit by direct substitution.

    Substitute x=3x=3 into the expression:
    Numerator: 32−9=9−9=03^2 - 9 = 9 - 9 = 0
    Denominator: 32+2(3)−3=9+6−3=15−3=123^2 + 2(3) - 3 = 9 + 6 - 3 = 15 - 3 = 12

  2. Evaluate the limit.

    Since the numerator approaches 0 and the denominator approaches a non-zero number (12), the limit of the fraction is 0.

    lim⁡x→3x2−9x2+2x−3=012=0\lim_{x \to 3} \frac{x^2 - 9}{x^2 + 2x - 3} = \frac{0}{12} = 0
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