Evaluate each limit. lim x → π cos 2 ( x ) + 3 cos ( x ) + 2 cos ( x ) + 1 {\displaystyle\lim_{x\to\pi}\frac{\cos^2\left(x\right)+3\cos\left(x\right)+2}{\cos^{}\left(x\right)+1}}

Welcome back, everyone. Determine the limit. Limit as X approaches to pi of two cosine square x minus 5 cosine of X plus 3 divided by cosine of X minus 1, and we're given 4 answer choices A says 0, B1, C -1, and D-2. So let's rewrite the limit. It says limit as X approaches to pi. In the numerator, we have to cosine squared of X minus 5 cosine of X plus 3. In the denominator we have cosine of x minus 1. And we begin with direct substitution. This gives us 2. Cosine squared of 2 pi. -5 cosine. Off to pine. Plus 3 in the denominator, we end up with cosine of 2 pi. -1 And now we can notice that this is an indeterminate form, we end up with 0 divided by 0. Now, what we want to do is potentially eliminate a possible hole, right? So we can notice that the numerator. Corresponds to a quadratic. equation if we We write it as follows. We can say that Z is equal to cosine of X, and then we end up with 2 z squared minus 5 Z. Plus 3. And now we can see that this is a parabola right in terms of Z and we can factor it out because we have a quadratic quinomial. And factoring it out, we end up with 2. See -3 multiplied by Z minus 1. And since he is cosine, well, we can substitute cosine back, right? But now considering our limit, if we go back to the limit itself, let's rewrite the numerator in its factored form. We get 2 cosine of x minus 3 multiplied by cosine of x minus 1 divided by cosine of X minus 1, and we indeed have a whole. Because there is a common factor in the numerator and the denominator of the rational function. And now we can apply the direct substitution for the remainder of the function, so we get to cosine. Of Ty. Minus 3. Now cosine of 2 pi is equal to 1, so we simply have 2 minus 3, which is -1, and that's our final answer which corresponds to the answer choice C. Thank you for watching.

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1. Limits and Continuity

Finding Limits Algebraically

Calculus

1. Limits and Continuity

Finding Limits Algebraically

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3:38 minutes

Problem 2.6.51

Textbook Question

Evaluate each limit.

$\lim_{x \to π} \frac{\cos^{2} (x) + 3 \cos (x) + 2}{\cos^{} (x) + 1}$

Verified step by step guidance

Identify the limit expression: $ \lim_{x \to \pi} \frac{\cos^2(x) + 3\cos(x) + 2}{\cos(x) + 1} $.

Substitute $ x = \pi $ into the expression to check if it results in an indeterminate form.

Notice that $ \cos(\pi) = -1 $, so substitute $ \cos(x) = -1 $ into the numerator and denominator.

Simplify the numerator: $ \cos^2(x) + 3\cos(x) + 2 = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0 $.

Simplify the denominator: $ \cos(x) + 1 = -1 + 1 = 0 $, indicating a $ \frac{0}{0} $ indeterminate form, so apply L'Hôpital's Rule or factor the expression to resolve the limit.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It helps in understanding how functions behave near points of interest, including points where they may not be defined. Evaluating limits is crucial for determining continuity, derivatives, and integrals.

Trigonometric Functions

Trigonometric functions, such as cosine, are periodic functions that relate angles to ratios of sides in right triangles. In this limit problem, the cosine function is evaluated at the point x = π, which is essential for finding the limit. Understanding the properties and values of trigonometric functions at specific angles is key to solving such problems.

Indeterminate Forms

Indeterminate forms occur in calculus when direct substitution in a limit leads to expressions like 0/0 or ∞/∞. These forms require further analysis, often using algebraic manipulation or L'Hôpital's rule, to resolve. Recognizing when a limit results in an indeterminate form is crucial for applying the appropriate techniques to evaluate it.

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