How do you prove that the limit of $\sqrt{x} = 3$ $sqrtx = 3$ as x approaches 9 using the epsilon delta proof?

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How do you prove that the limit of $\sqrt{x} = 3$ $sqrtx = 3$ as x approaches 9 using the epsilon delta proof?

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Answer 1 · 2017-10-22T00:15:53

Start from:

$| x - 9 | = ∣ ∣ \sqrt{x} - 3 ∣ ∣ ∣ ∣ \sqrt{x} + 3 ∣ ∣$ $abs(x-9) = abs(sqrtx-3)abs(sqrtx+3)$

for $x > 0$ $x > 0$ , so:

$∣ ∣ \sqrt{x} - 3 ∣ ∣ = \frac{| x - 9 |}{\sqrt{x} + 3}$ $abs(sqrtx-3) = abs(x-9)/(sqrtx+3)$

For any number $ε > 0$ $epsilon > 0$ choose now $δ_{ε} < 3 ε$ $delta_epsilon < 3epsilon$ .
For $x \in (9 - δ_{ε}, 9 + δ_{ε})$ $x in (9-delta_epsilon,9+delta_epsilon)$ we have;

$| x - 9 | < δ_{ε}$ $abs(x-9) < delta_epsilon$

and, as $\sqrt{x}$ $sqrtx$ is positive:

$\sqrt{x} + 3 > 3$ $sqrtx +3 > 3$

so:

$∣ ∣ \sqrt{x} - 3 ∣ ∣ = \frac{| x - 9 |}{\sqrt{x} + 3} < \frac{δ_{ε}}{3} < \frac{3 ε}{3} = ε$ $abs(sqrtx-3) = abs(x-9)/(sqrtx+3) < delta_epsilon/3 < (3epsilon)/3 = epsilon$

Thus:

$x \in (9 - δ_{ε}, 9 + δ_{ε}) \Rightarrow ∣ ∣ \sqrt{x} - 3 ∣ ∣ < ε$ $x in (9-delta_epsilon,9+delta_epsilon) => abs(sqrtx-3) < epsilon$

which proves that:

$lim_(x->9) sqrtx = 3$

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How do you prove that the limit of $\sqrt{x} = 3$ $sqrtx = 3$ as x approaches 9 using the epsilon delta proof?

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