Solved: "Use the precise of a limit to prove..."

CMIIh

Answered question

2021-10-15

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

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

Jaylen Fountain

Skilled2021-10-16Added 169 answers

Given:
To prove

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

Left hand limit

lim_{x ∖ \to 4^{-}} \frac{x - 4}{\sqrt{x} - 2} = lim_{h \to 0} \frac{4 - h - 4}{\sqrt{4 - h} - 2}

= lim_{h \to 0} \frac{- h}{\sqrt{4 - h} - 2} \times \frac{\sqrt{4 - h} + 2}{\sqrt{4 - h} + 2}

= lim_{h \to 0} \frac{(- h) (\sqrt{4 - h} + 2)}{{(\sqrt{4 - h})}^{2} - 2^{2}}

= lim_{h \to 0} \frac{- h (\sqrt{4 - h} + 2} {- h}}{}

= lim_{h \to 0} \sqrt{4 - h} + 2

= \sqrt{4} + 2

= 2 + 2

= 4

Right hand limit

lim_{x \to 4^{+}} \frac{x - 4}{\sqrt{x} - 2} = lim_{h \to 0} \frac{4 + h - 4}{\sqrt{4 + h} - 2}

= lim_{h \to 0} \frac{h}{\sqrt{4 + h} - 2} \times \frac{\sqrt{4 + h} + 2}{\sqrt{4 + h} + 2}

= lim_{h \to 0} \frac{(h) (\sqrt{4 + h} + 2)}{{(\sqrt{4 + h})}^{2} - 2^{2}}

= lim_{h \to 0} \frac{h (\sqrt{4 + h} + 2)}{h}

= \sqrt{4} + 2

= 4

Therefore the limit exists as left hand limit is equal to right hand limit
therefore

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

Jeffrey Jordon

Expert2022-08-30Added 2605 answers

Answer is given below (on video)

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