Problem 52 - Calculus: Early Transcendentals

Question

Evaluate limit. $$\lim _{x \rightarrow 3 \pi / 2} \frac{\sin ^{2} x+6 \sin x+5}{\sin ^{2} x-1}$$

Step-by-Step Solution

Answer

Question: Determine the limit of the function \(f(x)=\frac{\sin ^{2} x+6 \sin x+5}{\sin ^{2} x-1}\) as \(x\) approaches \(3\pi/2\). Answer: The limit of the function as \(x\) approaches \(3\pi/2\) is -2.

Step 1: Simplify the given function

Factorize the numerator and the denominator: $$\frac{\sin ^{2} x+6 \sin x+5}{\sin ^{2} x-1} = \frac{(\sin x+1)(\sin x+5)}{(\sin x+1)(\sin x-1)}$$

Step 2: Cancel out common terms

Both the numerator and the denominator have a common term \((\sin x+1)\), so we can cancel that out: $$\frac{(\sin x+1)(\sin x+5)}{(\sin x+1)(\sin x-1)} = \frac{\sin x+5}{\sin x-1}$$

Step 3: Substitute the limit value

Now that we have a simplified function, we can substitute \(x=3\pi/2\) and see if we get a determinate or indeterminate form: $$\lim _{x \rightarrow 3 \pi / 2} \frac{\sin x+5}{\sin x-1} = \frac{\sin \left(\frac{3 \pi}{2}\right)+5}{\sin \left(\frac{3 \pi}{2}\right)-1}$$ Since, \(\sin \left(\frac{3\pi}{2}\right)= -1\), then we have: $$\lim _{x \rightarrow 3 \pi / 2} \frac{\sin x+5}{\sin x-1} = \frac{-1+5}{-1-1} = \frac{4}{-2}$$

Step 4: Calculate the limit

Finally, we get the value of the limit by dividing -4 by 2: $$\lim _{x \rightarrow 3 \pi / 2} \frac{\sin ^{2} x+6 \sin x+5}{\sin ^{2} x-1} = \frac{4}{-2} = -2$$ Therefore, the limit of the given function as \(x\) approaches \(3\pi/2\) is -2.

Key Concepts

Evaluating Limits

Understanding how to evaluate limits is crucial in calculus. It requires analyzing a function's behavior as the input approaches a certain value. Consider the expression \(\lim _{x \rightarrow c} f(x)\), where \(c\) is the value that \(x\) is approaching. The objective is to find the value of \(f(x)\) as \(x\) gets infinitely close to \(c\), if it exists.

Evaluating limits can involve direct substitution if the function is continuous at \(c\), but when direct substitution leads to an indeterminate form like \(0/0\), other techniques such as factoring, rationalization, or applying limit laws may be necessary. For example, in the given exercise, simplifying the rational expression allows for direct substitution after the indeterminacy is resolved.

Simplifying Rational Expressions

Simplifying rational expressions, like those in the example exercise, is often essential before evaluating limits. A rational expression is a ratio of two polynomials. To simplify, look for common factors in the numerator and the denominator that can be cancelled out.

Factoring both the numerator and the denominator is the first step, as it reveals any common terms. Once the common factors are identified, they can be cancelled. This process simplifies the expression and can eliminate potential indeterminate forms. Simplification may also involve expanding expressions, combining like terms, or dividing each term by a common factor. The goal is to transform the original complex function into a simpler one that can be more easily evaluated at the limit value.

Trigonometric Limits

Limits involving trigonometric functions can present unique challenges. To evaluate trigonometric limits, like \(\lim _{x \rightarrow c} \sin x\) or \(\lim _{x \rightarrow c} \cos x\), it's important to know the values of these functions at key angles and to understand the trigonometric identities.

For example, if a limit leads to an expression like \(\sin c/\cos c\), recognizing this as \(\tan c\) can be helpful. In the provided exercise, knowing that \(\sin (3\pi/2) = -1\) allows us to substitute and find the limit. Moreover, understanding how to manipulate trigonometric functions can simplify the expressions before taking the limit, as shown by factoring in the exercise example.

Limit Laws

Limit laws are a set of rules that provide us methodologies to compute the limits of functions systematically. Some of the basic limit laws include:

• The sum law: \(\lim _{x \rightarrow c} (f(x)+g(x)) = \lim _{x \rightarrow c}f(x) + \lim _{x \rightarrow c}g(x)\)
• The product law: \(\lim _{x \rightarrow c} (f(x) \cdot g(x)) = \lim _{x \rightarrow c}f(x) \cdot \lim _{x \rightarrow c}g(x)\)
• The quotient law: Assuming \(\lim _{x \rightarrow c}g(x) eq 0\), \(\lim _{x \rightarrow c} (f(x)/g(x)) = \lim _{x \rightarrow c}f(x) / \lim _{x \rightarrow c}g(x)\)

By applying these laws, we can break down complex limit problems into simpler parts. But caution is required: they can only be used when the limits on the right-hand side exist and are finite. In the exercise, these laws allow us to cancel common factors in the numerator and the denominator and directly evaluate the limit of the simplified expression.