Question

determine the following limits, using ∞ or -∞ when appropriate, or state that they do not exist. a. lim(x→7⁺) 1/(x - 7) b. lim(x→7⁻) 1/(x - 7) c. lim(x→7) 1/(x - 7) a. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim(x→7⁺) 1/(x - 7)=□ (simplify your answer.) b. the limit does not exist and is neither ∞ nor -∞.

Explanation:

Step1: Analyze right - hand limit

When $x\to7^{+}$, $x - 7\to0^{+}$. Let $t=x - 7$, as $x\to7^{+}$, $t\to0^{+}$. Then $\lim_{x\to7^{+}}\frac{1}{x - 7}=\lim_{t\to0^{+}}\frac{1}{t}=+\infty$.

Step2: Analyze left - hand limit

When $x\to7^{-}$, $x - 7\to0^{-}$. Let $t=x - 7$, as $x\to7^{-}$, $t\to0^{-}$. Then $\lim_{x\to7^{-}}\frac{1}{x - 7}=\lim_{t\to0^{-}}\frac{1}{t}=-\infty$.

Step3: Analyze overall limit

Since $\lim_{x\to7^{-}}\frac{1}{x - 7}
eq\lim_{x\to7^{+}}\frac{1}{x - 7}$, the two - sided limit $\lim_{x\to7}\frac{1}{x - 7}$ does not exist.

Answer:

a. $\lim_{x\to7^{+}}\frac{1}{x - 7}=+\infty$
b. $\lim_{x\to7^{-}}\frac{1}{x - 7}=-\infty$
c. The limit $\lim_{x\to7}\frac{1}{x - 7}$ does not exist.