Question

question
evaluate the limit: $limlimits_{x \to 9} \dfrac{2x - 18}{\sqrt{x + 16} - 5}$

answer attempt 1 out of 2

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Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x + 16}+5$.

$$\lim_{x ightarrow9}\frac{2x - 18}{\sqrt{x + 16}-5}\times\frac{\sqrt{x + 16}+5}{\sqrt{x + 16}+5}$$

Step2: Simplify the numerator and denominator

The denominator becomes $(\sqrt{x + 16})^2-5^2=x + 16 - 25=x - 9$ (using the difference of squares formula $(a - b)(a + b)=a^2 - b^2$).
The numerator is $2(x - 9)(\sqrt{x + 16}+5)$ (factoring out 2 from $2x - 18$ gives $2(x - 9)$).
So the expression simplifies to:

$$\lim_{x ightarrow9}\frac{2(x - 9)(\sqrt{x + 16}+5)}{x - 9}$$

Step3: Cancel out common factors

Cancel out the common factor $(x - 9)$ (since $x
ightarrow9$ but $x
eq9$, we can cancel it).
We get:

$$\lim_{x ightarrow9}2(\sqrt{x + 16}+5)$$

Step4: Substitute $x = 9$

Substitute $x = 9$ into the expression:
$$2(\sqrt{9 + 16}+5)=2(\sqrt{25}+5)=2(5 + 5)=2\times10 = 20$$

Answer:

20