Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

find the indicated limit or state that it does not exist. In many cases, you will want to do some algebra before trying to evaluate the limit. $$ \lim _{x \rightarrow 2} \frac{x^{2}+7 x+10}{x+2} $$

Step-by-Step Solution

Answer

The limit is 7.

Step 1: Recognize the Indeterminate Form

First, substitute \(x = 2\) into the expression \(\frac{x^2 + 7x + 10}{x + 2}\). This gives \(\frac{2^2 + 7 \cdot 2 + 10}{2 + 2}\), which simplifies to \(\frac{18}{4}\). This is not an indeterminate form, which means substitution should be possible directly.

Step 2: Factor the Numerator

Examine the numerator \(x^2 + 7x + 10\) and factor it. We look for two numbers that multiply to \(10\) (the constant term) and add to \(7\) (the linear coefficient). The numbers \(5\) and \(2\) satisfy these conditions. Thus, the factorization will be \((x + 5)(x + 2)\).

Step 3: Simplify the Expression

Substitute the factored form \((x + 5)(x + 2)\) into the given expression. This gives us \(\frac{(x + 5)(x + 2)}{x + 2}\). Now, cancel the common factor \(x + 2\) in the numerator and the denominator. We are left with \(x + 5\).

Step 4: Evaluate the Simplified Limit

Now, evaluate the simplified expression \(x + 5\) as \(x\) approaches \(2\). Substituting \(x = 2\), we find that \(2 + 5 = 7\). Thus, the limit exists and equals \(7\).

Key Concepts

Understanding Limits in Calculus

In calculus, the concept of a limit is fundamental. It describes the value that a function approaches as the input approaches a certain point. Limits help us understand the behavior of functions at points where they are not necessarily defined. For instance, limits can show the trends of a curve or tell you how it behaves as it approaches infinity. When we say \[ \lim_{x \rightarrow a} f(x) \]we mean that as \(x\) gets closer and closer to \(a\), the function \(f(x)\) gets closer and closer to some particular value. This is crucial in understanding how to work with functions that have discontinuities, points of undefined behavior, or points of change. When you're faced with a complicated expression, it's a good practice to first consider if substitution gets you to the answer—a method frequently employed with straightforward polynomial expressions.

Understanding limits also lays the groundwork for other critical calculus concepts like derivatives and integrals. So, mastering them early makes learning calculus much smoother! Remember, finding a limit often involves either direct substitution, if the function is continuous at that point, or more advanced techniques if it's not.

Algebraic Manipulation for Limit Evaluation

Algebraic manipulation is a useful skill when evaluating limits. Often expressions will need to be simplified before direct evaluation is possible. This involves using techniques like factoring, expanding, or canceling terms to make the function more accessible.

For the exercise on \[ \lim_{x \rightarrow 2} \frac{x^2 + 7x + 10}{x + 2} \] algebraic manipulation was key.

• The numerator \(x^2 + 7x + 10\) was factored into \((x + 5)(x + 2)\), which showed a common term with the denominator.
• Cancelling out the common term \(x + 2\) left us with a simpler expression \(x + 5\).

The simplified expression could then be easily evaluated using substitution as \(x\) approaches 2. This algebraic manipulation helped bypass any complexity and made the limit clear and straightforward to calculate as 7.

Being comfortable with algebraic skills not only helps with limits but also with solving equations and inequalities, which pop up consistently in calculus problems.

Dealing with Indeterminate Forms

Indeterminate forms are situations in calculus where the limit doesn't immediately yield a clear value. In these cases, direct substitution could lead to answers like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), which don't tell us anything valuable. They indicate that more work is needed to evaluate the limit.

In such cases:

• Algebraic manipulation like factoring or finding common terms can often resolve indeterminate forms.
• L'Hopital's Rule, which applies derivatives to find limits of indeterminate forms, can also be useful but is usually a more advanced technique.

In the provided exercise, checking for an indeterminate form was the first step. It turned out not to be one since substituting directly into the expression didn't result in \(\frac{0}{0}\) or any of the other peculiar forms. This allowed for easy verification of the solution.

Recognizing indeterminate forms early can save a lot of time since it prevents chasing a calculation that goes nowhere. Once identified, appropriate methods can be leveraged to resolve the issue and find a meaningful limit.