Problem 51
Question
In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 2} \frac{2-x}{x^{2}-4} $$
Step-by-Step Solution
Verified Answer
The limit is \(-\frac{1}{4}\).
1Step 1: Identify the Function for the Limit
We have the function \( f(x) = \frac{2-x}{x^2-4} \) and need to evaluate \( \lim _{x \rightarrow 2} \frac{2-x}{x^2-4} \). Notice that directly substituting \( x = 2 \) results in \( \frac{0}{0} \), an indeterminate form. To resolve this, we must simplify the expression.
2Step 2: Factor the Denominator
The denominator \( x^2 - 4 \) is a difference of squares, which can be factored as \( (x-2)(x+2) \). Thus, the expression becomes \( \frac{2-x}{(x-2)(x+2)} \).
3Step 3: Simplify the Expression
Observe that \( 2-x \) can be rewritten as \( -(x-2) \). This allows us to cancel out \( x-2 \) from the numerator and denominator: \( \frac{-(x-2)}{(x-2)(x+2)} = -\frac{1}{x+2} \) for \( x eq 2 \).
4Step 4: Evaluate the Limit of the Simplified Expression
Now, evaluate \( \lim_{x \rightarrow 2} -\frac{1}{x+2} \). Since the expression is no longer indeterminate, substitute \( x = 2 \) directly: \( -\frac{1}{2+2} = -\frac{1}{4} \).
Key Concepts
Indeterminate FormsDifference of SquaresSimplifying Rational ExpressionsFactoring in Calculus
Indeterminate Forms
In calculus, indeterminate forms often arise when evaluating limits at specific points. An indeterminate form is a mathematical expression that does not initially give a clear answer about the limit. Common examples include the forms \( \frac{0}{0} \), \( \infty - \infty \), and \( 0 \times \infty \). These forms require further analysis and manipulation to resolve.
In the given exercise, when we substitute \( x = 2 \) in the function \( \frac{2-x}{x^2-4} \), it results in \( \frac{0}{0} \). This is a classic indeterminate form. Thus, it's necessary to perform algebraic manipulations to simplify the expression and find the limit. Without these steps, the value of the limit remains undefined.
In the given exercise, when we substitute \( x = 2 \) in the function \( \frac{2-x}{x^2-4} \), it results in \( \frac{0}{0} \). This is a classic indeterminate form. Thus, it's necessary to perform algebraic manipulations to simplify the expression and find the limit. Without these steps, the value of the limit remains undefined.
Difference of Squares
The expression \( x^2 - 4 \) in our problem is a perfect example of a difference of squares. A difference of squares follows the general formula \( a^2 - b^2 = (a-b)(a+b) \). This technique is used because it helps to factorize and simplify expressions that might be challenging to handle otherwise.
In the denominator, \( x^2 - 4 \) is rewritten as \((x-2)(x+2)\). By recognizing this pattern, we can simplify complex expressions further, making it easier to evaluate limits. Identifying such patterns is a powerful tool in calculus as it frequently allows us to transform indeterminate forms into manageable expressions.
In the denominator, \( x^2 - 4 \) is rewritten as \((x-2)(x+2)\). By recognizing this pattern, we can simplify complex expressions further, making it easier to evaluate limits. Identifying such patterns is a powerful tool in calculus as it frequently allows us to transform indeterminate forms into manageable expressions.
Simplifying Rational Expressions
Simplifying rational expressions involves reducing fractions by canceling common factors in the numerator and denominator. In our problem, after factoring the denominator, we notice that \(2-x\) can be expressed as \(-(x-2)\). This allows us to cancel the \(x-2\) terms.
Thus, \( \frac{2-x}{(x-2)(x+2)} \) becomes \(-\frac{1}{x+2}\) for all \(x eq 2\). Once simplified, the limit expression becomes straightforward to evaluate, highlighting how simplification can remove the indeterminacy that initially existed. This step is essential for executing direct substitution to find limits.
Thus, \( \frac{2-x}{(x-2)(x+2)} \) becomes \(-\frac{1}{x+2}\) for all \(x eq 2\). Once simplified, the limit expression becomes straightforward to evaluate, highlighting how simplification can remove the indeterminacy that initially existed. This step is essential for executing direct substitution to find limits.
Factoring in Calculus
Factoring is a key strategy used in calculus to simplify expressions, particularly when dealing with polynomial functions. By factoring, we can transform expressions to identify removable discontinuities or to simply analyze the behavior of functions near particular points.
In the exercise given, factoring the difference of squares in the denominator is pivotal. It uncovers the common term with the numerator, which can be subsequently canceled, revealing the true nature of the rational expression. This technique is invaluable because it converts seemingly complicated problems into more manageable ones, allowing for the straightforward application of limit laws and direct substitution.
In the exercise given, factoring the difference of squares in the denominator is pivotal. It uncovers the common term with the numerator, which can be subsequently canceled, revealing the true nature of the rational expression. This technique is invaluable because it converts seemingly complicated problems into more manageable ones, allowing for the straightforward application of limit laws and direct substitution.
Other exercises in this chapter
Problem 49
In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3} $$
View solution Problem 50
In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 1} \frac{(x-1)^{2}}{x^{2}-1} $$
View solution Problem 52
In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow-4} \frac{x+4}{16-x^{2}} $$
View solution Problem 54
In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 1 / 2} \frac{1-x-2 x^{2}}{1-2 x} $$
View solution