Question

question
evaluate the limit: $limlimits_{x \to 13} \frac{\sqrt{x - 12} - 1}{-5x + 65}$

answer attempt 1 out of 2

dne
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Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x - 12}+1$.

$$\lim_{x ightarrow13}\frac{\sqrt{x - 12}-1}{-5x + 65}\times\frac{\sqrt{x - 12}+1}{\sqrt{x - 12}+1}$$

Using the difference of squares formula $(a - b)(a + b)=a^{2}-b^{2}$ in the numerator, we get:

$$\lim_{x ightarrow13}\frac{(\sqrt{x - 12})^{2}-1^{2}}{(-5x + 65)(\sqrt{x - 12}+1)}=\lim_{x ightarrow13}\frac{x - 12 - 1}{-5(x - 13)(\sqrt{x - 12}+1)}$$

Simplify the numerator:

$$\lim_{x ightarrow13}\frac{x - 13}{-5(x - 13)(\sqrt{x - 12}+1)}$$

Step2: Cancel out the common factor

Cancel out the common factor $(x - 13)$ in the numerator and the denominator (note that $x
ightarrow13$ but $x
eq13$ at the limit point, so we can cancel):

$$\lim_{x ightarrow13}\frac{1}{-5(\sqrt{x - 12}+1)}$$

Step3: Substitute $x = 13$

Substitute $x = 13$ into the expression:
$$\frac{1}{-5(\sqrt{13 - 12}+1)}=\frac{1}{-5(1 + 1)}$$
Simplify the denominator:
$$\frac{1}{-5\times2}=-\frac{1}{10}$$

Answer:

$-\frac{1}{10}$