/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 Assertion $(\mathbf{A}): \lim _{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 99

Assertion $(\mathbf{A}): \lim _{\mathrm{x} \rightarrow 0^{\circ}}\left(\mathrm{x}^{\mathrm{x}^{*}}-\mathrm{x}^{\mathrm{x}}\right)=-1$ Reason \((\mathbf{R}): \lim _{x \rightarrow 0^{\prime}} x^{x}(x-1)=-1\)

Short Answer

Expert verified
#tag_title# Short answer based on step-by-step solution: #tag_content# The reason provided, \(\lim_{x \rightarrow 0^+} (x^x)(x-1) = -1\), is not valid because the natural logarithm of the limit in step 4 results in an undefined value (\(\ln(-1)\)). Consequently, this reason does not directly support the assertion. The correct limit needs to be found using other techniques.

Step by step solution

01

Rewrite the assertion in terms of limits.

Since the assertion is given in limit notation, rewrite it like this: \(\lim _{x \rightarrow 0^+}\left(x^{x^*} - x^x\right) = -1\).
02

Rewrite the reason in terms of limits.

Rewrite the reason as a limit statement: \(\lim _{x \rightarrow 0^+} (x^x)(x-1) = -1\).
03

Check the indeterminate form.

In order to check the indeterminate form, let's analyze the functions \(x^{x^*}\) and \(x^x\). As x approaches \(0\) from the positive side, both base (x) and exponent (either \(x^*\) or x) approaches 0. The form \(0^0\) is an indeterminate form.
04

Apply L'H么pital's rule.

Since the indeterminate form is \(0^0\) in this case, we cannot directly apply L'H么pital's rule. However, we can use logarithmic properties and L'H么pital's rule to analyze the behavior of the limit. For this, we take the natural logarithm of the limit: \(\ln\left(\lim _{x \rightarrow 0^+} (x^x)(x-1)\right) = \ln(-1)\) However, the natural logarithm of \(-1\) is undefined, so the reason provided is not valid.
05

Analyze the correct form

To analyzethe correct form of the expression, we need to find the limit of the given expression. Since their individual limits are indeterminate, we will focus on the difference \(x^{x^*} - x^x\). Now, let's replace \(x^*\) by its definition: \(x^* = \lim_{n \rightarrow \infty} \frac{x^n}{n}\). Thus, \(\lim _{x \rightarrow 0^+} \left(x^{\lim_{n \rightarrow \infty} \frac{x^n}{n}} - x^x\right)\) Now, use properties of limits to find the correct limit: \(\lim_{n \rightarrow \infty} \lim _{x \rightarrow 0^+} \left(x^{\frac{x^n}{n}} - x^x\right)\)
06

Find the correct limit

The correct limit is yet to be found, but from our analysis, we can see that the reason provided in the exercise does not lead us to the assertion directly. Further analysis should involve exploring other techniques to find the correct limit.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Column - I (A) If $\lim _{x \rightarrow \infty}\left(\sqrt{\left(x^{2}-x-1\right)}-a x-b\right)=0\(, where \)a>0$, then there exists atleast one a and \(b\) for which point (a, \(2 b\) ) lies on the line (B) If $\lim _{x \rightarrow 0} \frac{\left(1+a^{3}\right)+8 e^{1 / x}}{1+\left(1-b^{3}\right) e^{1 / x}}=2$, then there exists atleast one \(a\) and \(b\) for which point \(\left(a, b^{3}\right)\) lies on the line (C) If $\lim _{\mathrm{x} \rightarrow \infty}\left(\sqrt{\left(\mathrm{x}^{4}-\mathrm{x}^{2}+1\right)}-\mathrm{ax}^{2}-\mathrm{b}\right)=0$, then there exists atleast one a and \(\mathrm{b}\) for which point \((\mathrm{a},-2 \mathrm{~b})\) lies on the line (D) If \(\lim _{x \rightarrow-a} \frac{x^{7}+a^{7}}{x+a}=7\), where \(a<0\), then there exists atleast one a for which point \((a, 2)\) lies on the line. Column - II (P) \(\mathrm{y}=-3\) (Q) \(3 x-2 y-5=0\) (R) \(15 x-2 y-13=0\) (S) \(\mathrm{y}=2\)

$\sum_{r=1}^{\infty} \frac{r^{3}+\left(r^{2}+1\right)^{2}}{\left(r^{4}+r^{2}+1\right)\left(r^{2}+r\right)}$ is equal to (A) \(3 / 2\) (B) 1 (C) 2 (D) infinite

The value of $\lim _{x \rightarrow 0} \frac{(\tan (\\{x\\}-1)) \sin \\{x\\}}{\\{x\\}(\\{x\\}-1)}\(, where \)\\{x\\}$ denotes the fractional part function, is (A) is 1 (B) is tan 1 (C) is \(\sin 1\) (D) is non-existent

The true statement(s) is / are (A) If \(\lim _{x \rightarrow c} \mathrm{f}(\mathrm{x})=0\), then there must exist a number \(\mathrm{d}\) such that \(\mathrm{f}(\mathrm{d})<0.001\) (B) \(\lim _{x \rightarrow c} f(x)=L\), is equivalent to $\lim _{x \rightarrow c}(f(x)-L)=0$. (C) \(\lim _{x \rightarrow a}(f(x)+g(x))\) may exist even if the limits $\lim _{x \rightarrow i}$ \(\left(\mathrm{f}(\mathrm{x})\right.\) and $\lim _{\mathrm{x} \rightarrow \mathrm{a}}(\mathrm{g}(\mathrm{x})$ do not exist. (D) If \(\lim _{x \rightarrow a} f(x)\) exists and $\lim _{x \rightarrow a}(f(x)+g(x))$ does not exist, then \(\lim _{x \rightarrow a} g(x)\) does not exist.

$\lim _{\mathrm{n} \rightarrow \infty}\left\\{\frac{7}{10}+\frac{29}{10^{2}}+\frac{133}{10^{3}}+\ldots . .+\frac{5^{\mathrm{n}}+2^{\mathrm{n}}}{10^{\mathrm{n}}}\right\\}$ equals (A) \(3 / 4\) (B) 2 (C) \(5 / 4\) (D) \(1 / 2\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.