Problem 6 - Calculus Early Transcendentals

Question

$$\text { Evaluate } \lim _{x \rightarrow 3^{-}} \frac{1}{x-3} \text { and } \lim _{x \rightarrow 3^{+}} \frac{1}{x-3}$$

Step-by-Step Solution

Answer

Answer: As x approaches 3 from the left (3-), the limit of the function is equal to negative infinity: \(\lim_{x \rightarrow 3^{-}} \frac{1}{x-3} = - \infty\). As x approaches 3 from the right (3+), the limit of the function is equal to positive infinity: \(\lim_{x \rightarrow 3^{+}} \frac{1}{x-3} = \infty\).

Step 1: Analyze the function

The given function is \(\frac{1}{x-3}\). As x approaches 3, we notice that the denominator approaches 0, which means the function is going to have some kind of infinity behavior. In order to find the exact limit values, we need to analyze the function as x approaches 3 from the left (3-) and from the right (3+).

Step 2: When x approaches 3 from the left (3-)

When x approaches 3 from the left (or 3-), it means we are looking for the behavior of the function when x is slightly less than 3 (e.g. 2.9, 2.99, 2.999, etc). In this scenario, the expression in the denominator (x-3) will be slightly negative, making the entire function negative. Since the denominator approaches 0, the function will go towards negative infinity. Therefore, we conclude that: $$\lim_{x \rightarrow 3^{-}} \frac{1}{x-3} = - \infty$$

Step 3: When x approaches 3 from the right (3+)

When x approaches 3 from the right (or 3+), it means we are looking for the behavior of the function when x is slightly more than 3 (e.g. 3.1, 3.01, 3.001, etc). In this scenario, the expression in the denominator (x-3) will be slightly positive, making the entire function positive. Since the denominator approaches 0, the function will go towards positive infinity. Therefore, we conclude that: $$\lim_{x \rightarrow 3^{+}} \frac{1}{x-3} = \infty$$

Step 4: Summary of the findings

We have found that as x approaches 3 from the left (3-), the limit of the given function is equal to negative infinity: $$\lim_{x \rightarrow 3^{-}} \frac{1}{x-3} = - \infty$$ While as x approaches 3 from the right (3+), the limit of the given function is equal to positive infinity: $$\lim_{x \rightarrow 3^{+}} \frac{1}{x-3} = \infty$$