Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![**Problem Statement:**
Solve the following expression:
\[ \lim_{{x \to 1}} \frac{\sqrt{x^2 + 3} - 2x}{x^2 - 1} \]
---
**Solution:**
To solve this limit, we can use algebraic manipulation to simplify the expression and possibly apply L'Hôpital's Rule if needed. Here's a step-by-step solution:
1. Considering the expression and directly substituting \( x = 1 \) yields an indeterminate form \(\frac{0}{0}\). Thus, we'll need to manipulate the expression.
2. Let's rewrite the expression:
\[
\lim_{{x \to 1}} \frac{\sqrt{x^2 + 3} - 2x}{x^2 - 1}
\]
3. Factor the denominator:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
So, the expression becomes:
\[
\lim_{{x \to 1}} \frac{\sqrt{x^2 + 3} - 2x}{(x - 1)(x + 1)}
\]
4. Notice that direct evaluation by substituting \( x = 1 \) still results in the \(\frac{0}{0}\). We can consider rationalizing the numerator by multiplying and dividing by the conjugate:
\[
\sqrt{x^2 + 3} + 2x
\]
5. Multiply numerator and denominator by this conjugate:
\[
\lim_{{x \to 1}} \frac{(\sqrt{x^2 + 3} - 2x)(\sqrt{x^2 + 3} + 2x)}{(x - 1)(x + 1)(\sqrt{x^2 + 3} + 2x)}
\]
The numerator simplifies to:
\[
(\sqrt{x^2 + 3})^2 - (2x)^2 = x^2 + 3 - 4x^2 = -3x^2 + 3
\]
Therefore, the limit expression is:
\[
\lim_{{x \to 1}} \frac{3(1 - x^2)}{(x - 1)(x + 1)(\sqrt{x^2 + 3} + 2x)}
\]
Rewrite:
\[
\lim](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a4a1727-156d-4ade-b12f-025bad03a8ba%2Faea81202-f739-4747-aea3-4a64057a5f5c%2F9juap6b_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Solve the following expression:
\[ \lim_{{x \to 1}} \frac{\sqrt{x^2 + 3} - 2x}{x^2 - 1} \]
---
**Solution:**
To solve this limit, we can use algebraic manipulation to simplify the expression and possibly apply L'Hôpital's Rule if needed. Here's a step-by-step solution:
1. Considering the expression and directly substituting \( x = 1 \) yields an indeterminate form \(\frac{0}{0}\). Thus, we'll need to manipulate the expression.
2. Let's rewrite the expression:
\[
\lim_{{x \to 1}} \frac{\sqrt{x^2 + 3} - 2x}{x^2 - 1}
\]
3. Factor the denominator:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
So, the expression becomes:
\[
\lim_{{x \to 1}} \frac{\sqrt{x^2 + 3} - 2x}{(x - 1)(x + 1)}
\]
4. Notice that direct evaluation by substituting \( x = 1 \) still results in the \(\frac{0}{0}\). We can consider rationalizing the numerator by multiplying and dividing by the conjugate:
\[
\sqrt{x^2 + 3} + 2x
\]
5. Multiply numerator and denominator by this conjugate:
\[
\lim_{{x \to 1}} \frac{(\sqrt{x^2 + 3} - 2x)(\sqrt{x^2 + 3} + 2x)}{(x - 1)(x + 1)(\sqrt{x^2 + 3} + 2x)}
\]
The numerator simplifies to:
\[
(\sqrt{x^2 + 3})^2 - (2x)^2 = x^2 + 3 - 4x^2 = -3x^2 + 3
\]
Therefore, the limit expression is:
\[
\lim_{{x \to 1}} \frac{3(1 - x^2)}{(x - 1)(x + 1)(\sqrt{x^2 + 3} + 2x)}
\]
Rewrite:
\[
\lim
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