Question

Given the function \(q=\left(v^{2}+v-56\right) /(v-7),(v \neq 7),\) find the left-side limit and the right-side limit of \(q\) as \(v\) approaches 7 . Can we conclude from these answers that q has a limit as \(v\) approaches \(7 ?\)

Short answer

The limit of \( q \) as \( v \to 7 \) is 15.

Step-by-step solution

Analyzing the Function

Observe the given function \( q = \frac{v^2 + v - 56}{v - 7} \). Notice that direct substitution of \( v = 7 \) is not possible since the denominator \( v - 7 \) becomes zero, which is undefined. Hence, we proceed by simplifying the function to find the limits.

Factoring the Numerator

Factor the numerator \( v^2 + v - 56 \). This looks to be a quadratic expression, which we can factor by finding numbers that multiply to -56 and add up to 1. These numbers are 8 and -7. Therefore, \( v^2 + v - 56 = (v + 8)(v - 7) \).

Simplifying the Expression

Now substitute the factor back into the function: \( q = \frac{(v + 8)(v - 7)}{v - 7} \). As long as \( v eq 7 \), we can cancel \( (v - 7) \) from the numerator and the denominator, simplifying to \( q = v + 8 \).

Evaluating the Left-side Limit

To find the left-side limit of \( q \) as \( v \to 7^- \), use the simpler expression \( q = v + 8 \). Substitute a value slightly less than 7 into \( v \), such as 6.999. As \( v \to 7^- \), \( q \to 7 + 8 = 15 \). Thus, \( \lim_{{v \to 7^-}} q = 15 \).

Evaluating the Right-side Limit

To find the right-side limit of \( q \) as \( v \to 7^+ \), again use the expression \( q = v + 8 \). Substitute a value slightly greater than 7 into \( v \), such as 7.001. As \( v \to 7^+ \), \( q \to 7 + 8 = 15 \). Thus, \( \lim_{{v \to 7^+}} q = 15 \).

Checking if Limits Agree

Since both the left-side limit and the right-side limit as \( v \to 7 \) are equal to 15, \( \lim_{{v \to 7}} q = 15 \) exists. Therefore, even though the initial form was undefined at 7, the limits indicate that \( q \) has a limit as \( v \) approaches 7.

Key concepts

Understanding Left-side Limits

When calculating the left-side limit, we are interested in what happens to a function as the variable approaches a specific point from the left, or from values that are less than that point. For instance, with our function \( q = \frac{v^2 + v - 56}{v - 7} \), we're curious about what's happening as \( v \) gets closer to 7 from numbers slightly less than 7, like 6.999.
This is what we call approaching 7 from the "right on the number line." What's most important about finding the left-side limit is that we use values infinitesimally smaller than the point of interest.
For our function, we simplify \( q \) to \( v + 8 \) when \( v eq 7 \). Thus, as \( v \to 7^- \), \( q \to 15 \).
This process of estimating helps us understand how a function behaves at certain points of interest by looking at it from one direction.

Understanding Right-side Limits

The right-side limit is another crucial concept in calculus limits. Here, we examine the behavior of the function as the variable nears a particular number from the right, meaning from values larger than the specific point.
In our function example, \( q = \frac{v^2 + v - 56}{v - 7} \), we concentrate on what happens as \( v \) approaches 7 from values slightly greater than 7, such as 7.001.
This means we are "sneaking up" on the 7 from the right side on the number line. By employing the simplified expression, \( q = v + 8 \), we find that as \( v \to 7^+ \), \( q \to 15 \).
The alignment of both the left-side limit and the right-side limit is vital since it ensures the overall limit exists: they both reach 15 as \( v \) approaches 7.

Simplifying Expressions for Easier Calculations

Simplifying complex expressions is a powerful tool in mathematics, especially when working with limits. Simplifying allows us to remove undefined conditions or complicated expressions to find limits efficiently.
In the given problem, the function \( q = \frac{v^2 + v - 56}{v - 7} \) was initially undefined at \( v = 7 \) because the denominator became zero. To address this, we simplified it by factoring the numerator, \( v^2 + v - 56 \), into \((v + 8)(v - 7)\).
By canceling the common factor of \(v - 7\) from numerator and denominator, we obtained the simplified expression \(q = v + 8\) when \(v eq 7\). Such simplification is essential as it allows us to work around points where the function might be initially undefined yet still ascertain the behavior and limits as \(v\) approaches specific points. This approach not only helps with evaluating limits but also offers a clearer intuition about the function's behavior near the points of interest.