Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{w \rightarrow-k} \frac{w^{2}+5 k w+4 k^{2}}{w^{2}+k w}, k \neq 0$$

Short answer

Question: Find the limit of the given function as \(w\) approaches \(-k\). Function: \(\frac{w^{2}+5 k w+4 k^{2}}{w^{2}+k w}\) Answer: -3

Step-by-step solution

Simplify the expression

We have the expression $$\frac{w^{2}+5 k w+4 k^{2}}{w^{2}+k w}$$ We will attempt to simplify this expression by factoring out common terms. For the numerator, we can factor as follows: $$w^{2}+5 k w+4 k^{2} = (w + k)(w + 4k)$$ The expression now becomes $$\frac{(w + k)(w + 4k)}{w^{2}+k w}$$ We can factor \(w\) out from the denominator: $$w^{2}+k w = w(w + k)$$ Now the expression is $$\frac{(w + k)(w + 4k)}{w(w + k)}$$

Cancel the common terms

We can cancel the common term \((w + k)\) from the numerator and the denominator: $$\frac{(w + k)(w + 4k)}{w(w + k)} = \frac{w + 4k}{w}$$

Find the limit as w approaches -k

Now we will attempt to find the limit as w approaches -k by substituting it into the simplified expression: $$\lim _{w \rightarrow -k} \frac{w + 4k}{w} = \frac{(-k) + 4k}{-k} = \frac{3k}{-k}$$ Since \(k \neq 0\), this expression simplifies to $$\lim _{w \rightarrow -k} \frac{w + 4k}{w} = \frac{3k}{-k} = -3$$ Therefore, the limit of the given function as \(w\) approaches \(-k\) is -3.

Key concepts

Limit Evaluation

Understanding how to evaluate the limit of a function is a foundational concept in calculus. The limit describes the behavior of a function as the input approaches a certain value. In the textbook exercise, the problem asks us to find the limit as the variable, in this case, 'w', approaches the opposite of a constant 'k'.

To solve this, you will replace 'w' in the function with '-k', but only after simplifying the original expression if possible. Simplifying often involves factoring and canceling out alike terms to make substitution possible. In this particular problem, before directly substituting '-k' into the function, simplification is crucial. It prevents a potentially undefined situation (like dividing by zero) and helps in finding the limit value easily, which turns out to be a simple constant, -3, in our example. The process of evaluating limits often requires these algebraic manipulations to reach a point where the limit can be directly computed.

Factoring Polynomials

Factoring polynomials is an algebraic process that involves expressing a polynomial as the product of its factors. Factors are polynomials of lower degree that multiply together to give the original polynomial. In many calculus problems, like our example with the limit of a function, factoring is used to simplify the expression before applying the limit.

Let's look at the numerator: \(w^2 + 5kw + 4k^2\). How do we factor this? We look for two numbers that multiply to give \(4k^2\) and add to give \(5k\). In this case, these two numbers are \(k\) and \(4k\). Hence, we can write the numerator as \((w + k)(w + 4k)\). By breaking down the polynomial into its factors, we can often reduce the complexity of the calculus operations needed, such as canceling out common terms in the numerator and denominator.

Simplifying Expressions

Simplifying expressions is a crucial skill in mathematics that involves reducing an expression to its most basic form without changing its value. In calculus, this is particularly important before computing limits, as it can transform complex, indeterminate forms into something more manageable.

The goal is to make the expressions less daunting and reveal a straightforward path to finding the limit. As we did in the solution, after factoring, we simplify the expression by canceling out common terms—a technique that is often rewarding, as it brings us closer to the actual value of the limit. Note how after canceling the common term \((w + k)\), we are left with \(\frac{w + 4k}{w}\), an expression that can be easily evaluated by plugging in the value that 'w' approaches.