Questions: Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim as x approaches -12 of (sqrt(x^2+25)-13)/(x+12)

Solution Steps

To evaluate the limit \(\lim _{x \rightarrow-12} \frac{\sqrt{x^{2}+25}-13}{x+12}\), we can use algebraic manipulation to simplify the expression. Specifically, we can multiply the numerator and the denominator by the conjugate of the numerator to eliminate the square root.

Given the inequality \(3x - 2 \leq f(x) \leq x^2 - 3x + 7\) for \(x \geq 0\), we can use the Squeeze Theorem to find \(\lim _{x \rightarrow 3} f(x)\). We first find the limits of the bounding functions as \(x\) approaches 3 and then use these to determine the limit of \(f(x)\).

To evaluate the limit \(\lim _{x \rightarrow -12} \frac{\sqrt{x^{2}+25}-13}{x+12}\), we multiply the numerator and the denominator by the conjugate of the numerator:

\[ \frac{\sqrt{x^{2}+25}-13}{x+12} \cdot \frac{\sqrt{x^{2}+25}+13}{\sqrt{x^{2}+25}+13} = \frac{(\sqrt{x^{2}+25})^2 - 13^2}{(x+12)(\sqrt{x^{2}+25}+13)} \]

This simplifies to:

\[ \frac{x^2 + 25 - 169}{(x+12)(\sqrt{x^{2}+25}+13)} = \frac{x^2 - 144}{(x+12)(\sqrt{x^{2}+25}+13)} = \frac{(x-12)(x+12)}{(x+12)(\sqrt{x^{2}+25}+13)} \]

Canceling the common factor \((x+12)\):

\[ \frac{x-12}{\sqrt{x^{2}+25}+13} \]

Now, we can directly substitute \(x = -12\):

\[ \lim _{x \rightarrow -12} \frac{x-12}{\sqrt{x^{2}+25}+13} = \frac{-12-12}{\sqrt{(-12)^{2}+25}+13} = \frac{-24}{\sqrt{144+25}+13} = \frac{-24}{\sqrt{169}+13} = \frac{-24}{13+13} = \frac{-24}{26} = -\frac{12}{13} \]

Given \(3x - 2 \leq f(x) \leq x^2 - 3x + 7\) for \(x \geq 0\), we find the limits of the bounding functions as \(x\) approaches 3:

\[ \lim _{x \rightarrow 3} (3x - 2) = 3(3) - 2 = 9 - 2 = 7 \]

\[ \lim _{x \rightarrow 3} (x^2 - 3x + 7) = 3^2 - 3(3) + 7 = 9 - 9 + 7 = 7 \]

Since both limits are equal, by the Squeeze Theorem:

\[ \lim _{x \rightarrow 3} f(x) = 7 \]

Final Answer