Question

hw 2: problem 8
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0.25 $\frac{1}{4}$
the answer above is not correct.
evaluate the limit, if it exists. if not, enter dne.
$lim_{x \to 8} \frac{sqrt{x + 1} - 3}{x - 8} =$ 1/4
limits
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Explanation:

Step1: Rationalize the numerator

Multiply numerator and denominator by the conjugate $\sqrt{x+1}+3$:
$$\lim_{x \to 8} \frac{(\sqrt{x+1}-3)(\sqrt{x+1}+3)}{(x-8)(\sqrt{x+1}+3)}$$

Step2: Simplify numerator using difference of squares

Apply $(a-b)(a+b)=a^2-b^2$:
$$\lim_{x \to 8} \frac{(x+1)-9}{(x-8)(\sqrt{x+1}+3)} = \lim_{x \to 8} \frac{x-8}{(x-8)(\sqrt{x+1}+3)}$$

Step3: Cancel common factor $x-8$

Eliminate $(x-8)$ from numerator and denominator:
$$\lim_{x \to 8} \frac{1}{\sqrt{x+1}+3}$$

Step4: Substitute $x=8$ into the expression

Calculate the value of the simplified limit:
$$\frac{1}{\sqrt{8+1}+3} = \frac{1}{3+3}$$

Answer:

$\frac{1}{6}$