Key to "I have to compute this limit:limx→∞3(xx−3−x+2)"

Guadalupe Glass

Answered question

2022-03-31

I have to compute this limit:

lim_{x \to \infty} 3 (\sqrt{x} \sqrt{x - 3} - x + 2)

Luciana Cline

Beginner2022-04-01Added 14 answers

The two standard techniques work: multiply and divide by the conjugate, then divide both numerator and denominator by the highest power of x

lim_{x \to \infty} 3 (\sqrt{x} \sqrt{x - 3} - x + 2) = 3 lim_{x \to \infty} (\sqrt{x^{2} - 3 x} - (x - 2))

= 3 lim_{x \to \infty} \frac{(\sqrt{x^{2} - 3 x} - (x - 2)) (\sqrt{x^{2} - 3 x} + (x - 2))}{\sqrt{x^{2} - 3 x} + (x - 2)}

= 3 lim_{x \to \infty} \frac{(x^{2} - 3 x) - {(x - 2)}^{2}}{\sqrt{x^{2} - 3 x} + (x - 2)}

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

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

= 3 lim_{x \to \infty} \frac{\frac{1}{x} (x - 4)}{\frac{1}{x} (\sqrt{x^{2} - 3 x} + (x - 2))}

= 3 lim_{x \to \infty} \frac{1 - \frac{4}{x}}{\sqrt{1 - \frac{3}{x}} + 1 - \frac{2}{x}}}

= 3 (\frac{1}{\sqrt{1} + 1}) = \frac{3}{2}

Jeffrey Jordon

Expert2022-07-05Added 2605 answers

Answer is given below (on video)

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