Skip to main content
Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 2 - LimitsProblem 2.6.55
Chapter 2, Problem 2.6.55

Evaluate each limit. 


limx→π2sin(x)−1sin(x)−1{\(\displaystyle\[\lim\)_{x\(\to\]\frac{\pi}{2}\)}{\(\frac{\sin\left(x\right)-1}{\sqrt{\sin\left(x\right)}\)-1}}}

Verified step by step guidance
1
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:
3m

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.
Recommended video:
05:50
One-Sided Limits

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.
Recommended video:
5:50
Power Rules

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.
Recommended video:
05:34
Intro to Continuity
Related Practice
Textbook Question

b. Estimate a solution to the equation in the given interval using a root finder.


x^5+7x+5=0; (−1,0)

Textbook Question

Use the precise definition of infinite limits to prove the following limits.


limx→41(x−4)2=∞{\(\displaystyle\)\(\lim\)_{x\(\to\)4}}\(\frac{1}{\left(x-4\right)^2}\)=\(\infty\)

Textbook Question

Determine the interval(s) on which the following functions are continuous. 

f(x)=x^5+6x+17 / x^2−9

9
views
Textbook Question

Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions.

f(x)=x^2−3x+2 / x^10−x^9

Textbook Question

Sketch a possible graph of a function g, together with vertical asymptotes, satisfying all the following conditions.


g(2) =1,g(5) =−1,lim x→4 g(x) =−∞,lim x→7^− g(x) =∞,lim x→7^+ g(x) =−∞

Textbook Question

A sequence is an infinite, ordered list of numbers that is often defined by a function. For example, the sequence {2,4,6,8,…} is specified by the function f(n) = 2n, where n=1,2,3,….The limit of such a sequence is lim n→∞ f(n), provided the limit exists. All the limit laws for limits at infinity may be applied to limits of sequences. Find the limit of the following sequences or state that the limit does not exist. 


{0,1/2,2/3,3/4,…}, which is defined by f(n) = (n−1) / n, for n=1,2,3,…