Question

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

Short answer

The limit of the given trigonometric expression as x approaches 3π/2 is -2.

Step-by-step solution

Factor the numerator and denominator

To start, let's try to factor the numerator and denominator of the given expression. The numerator is: $$\sin^2 x + 6\sin x + 5$$ We can rewrite this as a quadratic expression in terms of sin(x): $$(\sin x + 1)(\sin x + 5)$$ The denominator is: $$\sin^2 x - 1$$ This is a difference of squares and can be factored as: $$(\sin x + 1)(\sin x - 1)$$ Now we have the factored expression: $$\frac{(\sin x + 1)(\sin x + 5)}{(\sin x + 1)(\sin x - 1)}$$

Simplify the expression

Now, we can cancel out the common \((\sin x + 1)\) factor in the numerator and the denominator. This leaves us with: $$\frac{\sin x + 5}{\sin x - 1}$$

Apply the limit

Now we can directly apply the limit to the simplified expression, as follows: $$\lim _{x \rightarrow 3\pi/2} \frac{\sin x + 5}{\sin x - 1}$$ Evaluating the sine functions at x = 3π/2: $$\sin(3\pi/2)=-1$$ Plugging this value into the limit, we get: $$\lim _{x \rightarrow 3\pi/2} \frac{\sin x + 5}{\sin x - 1} = \frac{-1+5}{-1-1} = \frac{4}{-2}$$

Final answer

So the limit is: $$\lim _{x \rightarrow 3\pi/2} \frac{\sin ^{2} x+6 \sin x+5}{\sin ^{2} x-1} = -2$$

Key concepts

Limit Evaluation

Evaluating limits is a fundamental concept in calculus, focusing on the behavior of functions as the input approaches a particular point. To find the limit of a function, we consider what value the function approaches as the input gets infinitely close to a certain value, rather than the actual value of the function at that point.
When we encounter expressions involving limits, especially those that lead to zero in the denominator, we often need to do some algebraic manipulation, such as factoring, to evaluate the limit correctly. This lets us simplify complex expressions and eliminate indeterminate forms.
The goal is to make it possible to easily substitute the point into the simplified expression without dealing with undefined results. Mastery of limit evaluation paves the way for more advanced topics in calculus, such as derivatives and integrals.

• Understand the behavior of the function near the point of interest.
• Simplify complex expressions through factoring or other algebraic techniques.
• Substitute the point into the simplified expression to find the limit value.

Trigonometric Limits

Trigonometric limits involve expressions containing sine, cosine, or other trigonometric functions. These types of problems test your understanding of both limits and trigonometry, combining the two to evaluate limits where trigonometric functions are involved. Typically, they arise from manipulating trigonometric identities or simplifying expressions before evaluating the limit.
In the example given, the trigonometric function \(\sin x\) is part of the expression. Often, special attention is required when approaching points where these functions have significant behavior changes—like near asymptotes or special angles such as \(\pi\) or \(\pi/2\). In many trigonometric limit problems, recognizing patterns and using known trigonometric identities can simplify the process.
When the value is substituted directly and results in an indeterminate form (like 0/0), factoring may further simplify the expression, enabling you to apply the direct substitution method efficiently.

• Identify and use trigonometric identities to simplify expressions.
• Avoid indeterminate forms through algebraic simplification.
• Ensure proper substitution once simplifications are complete.

Factoring Expressions

Factoring is a crucial algebraic tool, especially when evaluating limits that involve rational expressions. Factoring simplifies complex expressions, making it possible to cancel out terms and eliminate indeterminate forms, which often appear when both the numerator and denominator approach zero.
In the exercise, the numerator and denominator were both factored. The numerator, \(\sin^2 x + 6\sin x + 5\), was expressed as \((\sin x + 1)(\sin x + 5)\). Similarly, the denominator, \(\sin^2 x - 1\), was factored into \((\sin x + 1)(\sin x - 1)\), using the difference of squares identity.
By factoring both expressions, common terms were canceled, greatly simplifying the limit evaluation process. This method helps turn seemingly difficult problems into manageable ones, focusing on canceling out and simplifying differences or sums with common terms.

• Look for common factors or use identities, such as difference of squares.
• Factor both numerator and denominator to simplify expressions.
• Cancel common factors to resolve indeterminate forms or simplify calculations.