Chapter 3: Q. 49 (page 311)

Calculate each of the limits in Exercises $49 - 64$ . Some of these limits are made easier by considering the logarithm of then limit first, and some are not.
$\lim_{x \to 0^{+}} x^{\ln x}$ .

Short Answer

Expert verified

The exact value of the limit $\lim_{x \to 0^{+}} x^{\ln x}$ is, $\infty$ .

Step by step solution

Step 1. Given information

$\lim_{x \to 0^{+}} x^{\ln x}$ .

Step 2. Taking logarithm of the limit.

$\lim_{x \to 0^{+}} \ln (x^{\ln x}) = \lim_{x \to 0^{+}} (\ln x) \ln (x) = \lim_{x \to 0^{+}} (\ln x)^{2}$

The limit using L'Hopital's rule is given below:

$\lim_{x \to 0^{+}} (\ln x)^{2} = \lim_{x \to 0^{+}} \frac{(\ln x)^{3}}{\ln x}$ [ in the form of $\frac{0}{0}$ ]

$= \lim_{x \to 0^{+}} \frac{3 (\ln x)^{2} \cdot \frac{1}{x}}{\frac{1}{x}}$ [ Using L'Hopital's rule]

$= \lim_{x \to 0^{+}} 3 (\ln x)^{2} \cdot \frac{1}{x^{2}} = 3 \lim_{x \to 0^{+}} \frac{(\ln x)^{2}}{x^{2}}$

$= 3 \lim_{x \to 0^{+}} \frac{2 (\ln x) \cdot \frac{1}{x}}{2 x}$ [Using L'Hopital's rule]
$= 3 \lim_{x \to 0^{+}} \frac{(\ln x)}{x^{2}}$

$= 3 \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{2 x}$ [L'Hopital's rule]

$= 3 \lim_{x \to 0^{+}} \frac{1}{2 x^{2}} = \frac{3}{2} \cdot \frac{1}{0} = \infty$

Step 3. Therefore, the value of the limit is given by,

$\lim_{x \to 0^{+}} \ln (x^{\ln x}) = e^{\infty} = \infty$

Therefore, the exact value of the limit is, $\infty$ .

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.

Calculate each of the limits in Exercises $49 - 64$ . Some of these limits are made easier by considering the logarithm of then limit first, and some are not.
$\lim_{x \to 0^{+}} x^{\ln x}$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. Taking logarithm of the limit.

Step 3. Therefore, the value of the limit is given by,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Statistics

Geometry

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.

Calculate each of the limits in Exercises 49-64. Some of these limits are made easier by considering the logarithm of then limit first, and some are not.limx→0+xlnx.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Taking logarithm of the limit.

Step 3. Therefore, the value of the limit is given by,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Statistics

Geometry

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.

Calculate each of the limits in Exercises $49 - 64$ . Some of these limits are made easier by considering the logarithm of then limit first, and some are not.
$\lim_{x \to 0^{+}} x^{\ln x}$ .