Question

question
evaluate the limit: $limlimits_{x \to 81} \frac{81 - x}{9 - \sqrt{x}}$

answer attempt 1 out of 2

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dne

Explanation:

Step1: Factor the numerator

Notice that \(81 - x\) is a difference of squares, so \(81 - x=(9 - \sqrt{x})(9+\sqrt{x})\) since \(x = (\sqrt{x})^2\) and \(a^2 - b^2=(a - b)(a + b)\) with \(a = 9\) and \(b=\sqrt{x}\).
So the function becomes \(\frac{(9 - \sqrt{x})(9+\sqrt{x})}{9 - \sqrt{x}}\) for \(x
eq81\) (we can cancel the non - zero factor \(9-\sqrt{x}\) when \(x
eq81\)).

Step2: Cancel the common factor

Cancel out the common factor \(9 - \sqrt{x}\) from the numerator and the denominator. We get the simplified function \(y = 9+\sqrt{x}\) for \(x
eq81\).

Step3: Evaluate the limit

Now, we find the limit as \(x
ightarrow81\) of \(9+\sqrt{x}\). Using the limit laws (the limit of a sum is the sum of the limits and the limit of a square root function \(\lim_{x
ightarrow a}\sqrt{x}=\sqrt{a}\) for \(a\geq0\)), we have \(\lim_{x
ightarrow81}(9+\sqrt{x})=\lim_{x
ightarrow81}9+\lim_{x
ightarrow81}\sqrt{x}\).
Since \(\lim_{x
ightarrow81}9 = 9\) and \(\lim_{x
ightarrow81}\sqrt{x}=\sqrt{81}=9\), then \(\lim_{x
ightarrow81}(9+\sqrt{x})=9 + 9=18\).

Answer:

\(18\)