Chapter 3: Q. 32 (page 314)

L’Hˆopital’s Rule limit calculations: Calculate each of the
limits that follow. Some of these limits are easier to calculate
by using L’Hopital’s rule, and some are not.
$\lim_{x \to 0^{+}} x^{1 - \cos x}$

Short Answer

Expert verified

Limit is $1$ and it is easy to find using L'Hospital rule

Step by step solution

Step 1. Given information

An expression is given as $\lim_{x \to 0^{+}} x^{1 - \cos x}$

Step 2. Evaluating limit

Use properties of limits as,

$\lim_{x \to 0^{+}} \ln x^{1 - \cos x} = \lim_{x \to 0^{+}} (1 - \cos x) \ln x = \lim_{x \to 0^{+}} \frac{\ln x}{(1 - \cos x)} = \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{- \frac{(- (- \sin x))}{(1 - \cos x)^{2}}} = - \lim_{x \to 0^{+}} \frac{(1 - \cos x)^{2}}{x \sin x} = - \lim_{x \to 0^{+}} \frac{1 - 2 \cos x + \cos^{2} x}{x \sin x} = - \lim_{x \to 0^{+}} \frac{2 \sin x + 2 \cos x (- \sin x)}{x \cos x + \sin x \times 1} = - 2 \lim_{x \to 0^{+}} \frac{\sin x - \sin x \cos x}{x \cos x + \sin x} = - 2 \lim_{x \to 0^{+}} \frac{\cos x - \sin x (- \sin x) - \cos x \cos x}{x (- \sin x) + \cos x + \cos x} = - 2 \lim_{x \to 0^{+}} \frac{\cos x - \sin^{2} x - \cos^{2} x}{- x \sin x + 2 \cos x} = - 2 \lim_{x \to 0^{+}} \frac{\cos x - (\sin^{2} x + \cos^{2} x)}{- x \sin x + 2 \cos x} = - 2 \lim_{x \to 0^{+}} \frac{\cos x - 1}{- x \sin x + 2 \cos x} = - 2 \frac{\cos 0 - 1}{- 0 \sin 0 + 2 \cos 0} = - 2 \times \frac{1 - 1}{0 + 2 \times 1} = - 2 \times \frac{0}{2} = 0 \underset{x \to 0^{+}}{\Rightarrow \lim} x^{1 - \cos x} = e^{0} = 1$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

L’Hˆopital’s Rule limit calculations: Calculate each of the
limits that follow. Some of these limits are easier to calculate
by using L’Hopital’s rule, and some are not.
$\lim_{x \to 0^{+}} x^{1 - \cos x}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Evaluating limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Pure Maths

Statistics

Mechanics Maths

Decision Maths

Geometry

Study anywhere. Anytime. Across all devices.

L’Hˆopital’s Rule limit calculations: Calculate each of thelimits that follow. Some of these limits are easier to calculateby using L’Hopital’s rule, and some are not.limx→0+x1-cosx

Short Answer

Step by step solution

Step 1. Given information

Step 2. Evaluating limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Pure Maths

Statistics

Mechanics Maths

Decision Maths

Geometry

Study anywhere. Anytime. Across all devices.

L’Hˆopital’s Rule limit calculations: Calculate each of the
limits that follow. Some of these limits are easier to calculate
by using L’Hopital’s rule, and some are not.
$\lim_{x \to 0^{+}} x^{1 - \cos x}$