Expert Answer to "Use the precise definition of a limit to..."

generals336

Answered question

2021-10-16

Use the precise definition of a limit to prove the following limits.

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = - 1

unessodopunsep

Skilled2021-10-17Added 105 answers

To prove

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = - 1

Considering LHS

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = lim_{x \to 3} \frac{x^{2} - 3 x - 4 x + 12}{x - 3}

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = lim_{x \to 3} \frac{x (x - 3) - 4 (x - 3)}{x - 3}

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = lim_{x \to 3} \frac{(x - 3) (x - 4)}{x - 3}

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = lim_{x \to 3} (x - 4)

Now putting limit x=3, in equation

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = (3 - 4)

lim_{x \to 3} \frac{x^{2} - 7 x + 12}{x - 3} = - 1

Hence proved

Jeffrey Jordon

Expert2022-08-30Added 2605 answers

Answer is given below (on video)

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