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2021-10-13

Evaluate the following using Limits of Theorem
$lim_{x \to 16} (\frac{x^{2} - 256}{4 - \sqrt{x}})$

Answer & Explanation

Clelioo

Skilled2021-10-14Added 88 answers

Using the property

a^{2} - b^{2} = (a - b) (a + b)

Consider

L = lim_{x \to 16} (\frac{x^{2} - 256}{4 - \sqrt{x}})

= lim_{x \to 16} (\frac{x^{2} - 16^{2}}{4 - \sqrt{x}})

= lim_{x \to 16} (\frac{(x - 16) (x + 16)}{4 - \sqrt{x}})

= lim_{x \to 16} (\frac{- (4^{2} - {(\sqrt{x})}^{2}) (x + 16)}{4 - \sqrt{x}})

= lim_{x \to 16} (\frac{- (4 - \sqrt{x}) (4 + \sqrt{x}) (x + 16)}{4 - \sqrt{x}})

= lim_{x \to 16} - (4 + \sqrt{x}) (x + 16)

= - (4 + 4) (16 + 16) = - 256

Hence

lim_{x \to 16} \frac{x^{2} - 256}{4 - \sqrt{x}} = - 256

Jeffrey Jordon

Expert2022-06-29Added 2605 answers

Answer is given below (on video)

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