Evaluate each limit. lim x → π 2 sin ( x ) − 1 sin ( x ) − 1 {\displaystyle\lim_{x\to\frac{\pi}{2}}{\frac{\sin\left(x\right)-1}{\sqrt{\sin\left(x\right)}-1}}}

Welcome back, everyone. Evaluate the limit as x approaches 0 of the expression, cosine x divided by square root of cosine x minus 1. We're given 4 answer choices AS 3, B2, C 1, and D 0. So let's look at this function and let's define the limit. Limit as X approaches 0, of the expression, cosine of x minus 1. Divided by square root of cosine x. -1 And as usual, let's apply the right substitution. We get cosine of 0, which is 1. -1 Divided by square root of cosine 0. Which is once gain 1 minus 1. So the numerator and the denominator do become 0. And we end up with an indeterminate form 0 divided by 0 because this is an indeterminate form we want to potentially eliminate a whole, right? And we can do that by noticing that we can apply a difference of squares factorization for the numerator. We can rewrite cosine x minus 1 as square root of cosine x squared. Minus 1 squad and now this is a difference of squares formula we can factorize it as Square root of cosine x the first term minus the second term, which is 1. And then multiplied by the first term. Square root of cosine x + 1. If we now substitute this factored expression into the limit. We end up with limit. Of square root of cosine x minus 1. Multiplied by square root of cosine x + 1. Divided by Square root of cosine x minus 1. And this eliminates a hole, right, because we have a common factor. Let's cross it out. And now we can apply direct substitution once again, which gives us square root of cosine of zero. Plus one Cosine of 0 is 1 and square root of 1 is 1, so we get 1 plus 1. Which is a finite number, that's 2. So we have determined that the correct answer to this problem is B2. Thank you for watching.

Table of contents

Skip topic navigation

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

1. Limits and Continuity

Finding Limits Algebraically

Calculus

1. Limits and Continuity

Finding Limits Algebraically

Previous problemNext problem

3:03 minutes

Problem 2.6.55

Textbook Question

Evaluate each limit.

$\lim_{x \to \frac{π}{2}} \frac{\sin (x) - 1}{\sqrt{\sin (x)} - 1}$

Verified step by step guidance

Identify the limit expression: $ \lim_{x \to \frac{\pi}{2}} \frac{\sin(x) - 1}{\sqrt{\sin(x)} - 1} $.

Substitute $ x = \frac{\pi}{2} $ into the expression to check for indeterminate form: $ \sin\left(\frac{\pi}{2}\right) = 1 $, leading to $ \frac{0}{0} $.

Apply L'Hôpital's Rule, which is used for limits of the form $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $, by differentiating the numerator and the denominator separately.

Differentiate the numerator: $ \frac{d}{dx}[\sin(x) - 1] = \cos(x) $.

Differentiate the denominator: $ \frac{d}{dx}[\sqrt{\sin(x)} - 1] = \frac{1}{2\sqrt{\sin(x)}} \cdot \cos(x) $.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

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

Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. In this question, we are evaluating the limit of a function as x approaches π/2, which is crucial for understanding the behavior of the function near that point.

L'Hôpital's Rule

L'Hôpital's Rule is a method used to evaluate limits that result in indeterminate forms, such as 0/0 or ∞/∞. When faced with such forms, we can differentiate the numerator and denominator separately and then take the limit again, which can simplify the evaluation process.

Continuity and Discontinuity

Continuity refers to a function being unbroken and having no gaps at a point, while discontinuity indicates a break or jump in the function's value. Understanding whether the function is continuous at x = π/2 helps determine if the limit can be directly evaluated or if further analysis is needed.

Watch next

Master Finding Limits by Direct Substitution with a bite sized video explanation from Patrick

Start learning