Question

Find the limits of \(q=(v+2)(v-3)\) (a) As \(v \rightarrow-1\) (b) As \(v \rightarrow 0\) (c) As \(v \rightarrow 5\)

Short answer

The limits are -4, -6, and 14, as \(v\) approaches -1, 0, and 5 respectively.

Step-by-step solution

Understanding the Concept of Limits

To find the limit of a function as a variable approaches a particular value, we substitute the value into the function if the function is continuous at that point. Limits examine the behavior of functions as they get close to a specific point.

Limit as \(v\rightarrow-1\)

Substitute \(v = -1\) into \(q = (v+2)(v-3)\). This gives \((-1+2)(-1-3) = (1)(-4) = -4\). Therefore, the limit of \(q\) as \(v\) approaches -1 is -4.

Limit as \(v\rightarrow0\)

Substitute \(v = 0\) into \(q = (v+2)(v-3)\). This gives \((0+2)(0-3) = (2)(-3) = -6\). Therefore, the limit of \(q\) as \(v\) approaches 0 is -6.

Limit as \(v\rightarrow5\)

Substitute \(v = 5\) into \(q = (v+2)(v-3)\). This gives \((5+2)(5-3) = (7)(2) = 14\). Therefore, the limit of \(q\) as \(v\) approaches 5 is 14.

Key concepts

Continuity

In calculus, continuity is an essential concept when dealing with functions and their limits. A function is said to be continuous at a point if there's no interruption or gap in its graph at that point. Simply put, you can draw the function at that point without lifting your pencil from the paper. Mathematically, a function \( f(x) \) is continuous at \( x = a \) if the following three conditions are satisfied:

• The function \( f(a) \) is defined.
• The limit of the function as \( x \) approaches \( a \), \( \lim_{{x \to a}} f(x) \), exists.
• The limit is equal to the function value at that point, meaning \( \lim_{{x \to a}} f(x) = f(a) \).

When we say a function is continuous over an interval, it means the function is continuous at every point in that interval. It's crucial for determining limits, because if a function is continuous at a point, finding its limit simply means evaluating the function at that point.

Substitution Method

The substitution method is a straightforward technique used to evaluate limits. When dealing with continuous functions, you can directly substitute the value that the variable approaches into the function to find the limit. For instance, if a function \( f(v) \) is continuous at \( v = a \), then \( \lim_{{v \to a}} f(v) = f(a) \).
This is exactly what we do in the original exercise. Consider the function \( q(v) = (v+2)(v-3) \):

• To find \( \lim_{{v \to -1}} q(v) \), we substitute \( v = -1 \), resulting in \( (-1+2)(-1-3) = 1 \times (-4) = -4 \).
• To find \( \lim_{{v \to 0}} q(v) \), we substitute \( v = 0 \), resulting in \( (0+2)(0-3) = 2 \times (-3) = -6 \).
• And for \( \lim_{{v \to 5}} q(v) \), substituting \( v = 5 \) gives us \( (5+2)(5-3) = 7 \times 2 = 14 \).

This method works nicely for functions where direct substitution does not lead to indeterminate forms like \( \frac{0}{0} \). If the function is not continuous at \( a \), or if substitution gives an indeterminate form, other methods such as factoring, rationalizing, or L'Hospital's rule may be needed.

Evaluating Functions at Points

Evaluating a function at a particular point means substituting a specific value for the variable in the function. This is similar to what you do in the substitution method when finding limits of continuous functions.
Consider the exercise with the function \( q(v) = (v+2)(v-3) \). By evaluating this function for different values of \( v \), we determine the behavior of the function around those points. For example, to find \( q(0) \), you would substitute \( v = 0 \) into the expression and simplify:

• \( q(0) = (0+2)(0-3) = 2 \times (-3) = -6 \).

Evaluating functions at points is also crucial in confirming the continuity of the function at that point. If the evaluated value matches the limit as you approach that particular point, then the function is continuous there. This practice strengthens understanding of the graphical behavior of functions and their continuous nature at various points in their domain.