Question

question
evaluate the limit: $limlimits_{x \to 8} \frac{\sqrt{x + 1}}{x + 12}$

answer attempt 1 out of 2

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dne

Explanation:

Step1: Substitute \( x = 8 \) into the function

Since the function \( \frac{\sqrt{x + 1}}{x + 12} \) is a rational function (a quotient of two functions, where the numerator is a square - root function and the denominator is a linear function) and the denominator \( x+12 \) is not zero when \( x = 8 \) (because when \( x = 8 \), \( x + 12=8 + 12=20
eq0 \)), we can use the direct substitution property of limits. The direct substitution property states that if \( f(x) \) is continuous at \( x=a \), then \( \lim_{x
ightarrow a}f(x)=f(a) \). The function \( y=\frac{\sqrt{x + 1}}{x + 12} \) is continuous at \( x = 8 \) because the numerator \( \sqrt{x + 1} \) is continuous for \( x\geq - 1 \) (since the square - root function \( \sqrt{u} \) is continuous for \( u\geq0 \), and here \( u=x + 1 \)) and the denominator \( x + 12 \) is a linear function, continuous everywhere, and non - zero at \( x = 8 \).

Substitute \( x = 8 \) into the numerator: \( \sqrt{8 + 1}=\sqrt{9}=3 \)

Substitute \( x = 8 \) into the denominator: \( 8+12 = 20 \)

Step2: Calculate the value of the fraction

Now we have the fraction \( \frac{3}{20} \) after substitution.

Answer:

\( \frac{3}{20} \)