/*! 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 36 Find the limits. $$\lim _{x \r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 36

Find the limits. $$\lim _{x \rightarrow 4} \frac{4 x-x^{2}}{2-\sqrt{x}}$$

Short Answer

Expert verified
The limit is indeterminate initially; rationalize the denominator and simplify.

Step by step solution

01

Direct Substitution

First, attempt to substitute the value of the limit directly into the expression: \( \lim_{x \to 4} \frac{4x - x^2}{2 - \sqrt{x}} \). By substituting \( x = 4 \), we get \( \frac{4(4) - 4^2}{2 - \sqrt{4}} \). This simplifies to \( \frac{16 - 16}{2 - 2} = \frac{0}{0} \), which is an indeterminate form. Therefore, we need to use algebraic manipulation to simplify the expression.
02

Rationalizing the Denominator

To eliminate the square root in the denominator, rationalize the expression by multiplying the numerator and the denominator by the conjugate of the denominator: \( 2 + \sqrt{x} \). Now the expression becomes \( \frac{4x - x^2}{2 - \sqrt{x}} \cdot \frac{2 + \sqrt{x}}{2 + \sqrt{x}} \).
03

Simplify the Expression

Multiply the numerators: \( (4x - x^2)(2 + \sqrt{x}) = 8x + 4x\sqrt{x} - 2x^2 - x^2\sqrt{x} \). Multiply the denominators using the difference of squares formula: \((2 - \sqrt{x})(2 + \sqrt{x}) = 4 - x \).
04

Further Simplify the Numerator

Rewrite the numerator to facilitate cancellation with the new denominator: \( 8x + 4x\sqrt{x} - 2x^2 - x^2\sqrt{x} \). Now observe common factors of \((x-4)\) or simplifying directly might be complex due to the expression's structure; consider expanding any factors if needed.
05

Substitute Limit Value Again

After simplifying the expression, substitute \( x = 4 \) into the revised expression to find the limit. Evaluate by carefully substituting or resolve any indeterminate forms that might still exist if previous steps were not clear in simplification.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
When evaluating limits in calculus, sometimes direct substitution of the limit value into the function results in expressions like \( \frac{0}{0} \). These expressions are called indeterminate forms. They do not give a clear answer, and thus need further manipulation.
  • An indeterminate form suggests more work is needed to find the actual limit.
  • These forms indicate a need for algebraic manipulation to simplify the expression.
  • Other indeterminate forms include \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and \( \infty - \infty \), among others.
In this particular problem, substituting \( x = 4 \) directly into \( \frac{4x - x^2}{2 - \sqrt{x}} \) yields \( \frac{0}{0} \). This tells us we need a different strategy to find the limit.
Rationalizing the Denominator
Rationalizing the denominator is a method used to eliminate square roots or other irrational numbers. In this exercise, the denominator \( 2 - \sqrt{x} \) needs to be adjusted.
  • We multiply the numerator and the denominator by the conjugate of the denominator, which is \( 2 + \sqrt{x} \).
  • This process uses the identity \((a-b)(a+b)=a^2-b^2\) to simplify expressions.
  • The goal is to convert the denominator into a simple expression without radicals.
Here, we multiply \( \frac{4x - x^2}{2 - \sqrt{x}} \) by \( \frac{2 + \sqrt{x}}{2 + \sqrt{x}} \) to simplify the denominator to \( 4 - x \). This is crucial for further simplifying the limit.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions to achieve desired forms. This may include factoring, expanding, or canceling terms.
  • In this problem, after rationalizing the denominator, simplify the numerator with the new denominator \( 4 - x \).
  • Multiply the expressions carefully using distribution.
  • Look for and cancel common factors to simplify.
In our exercise, after the numerators \((4x - x^2)(2 + \sqrt{x})\) and denominators \((2 - \sqrt{x})(2 + \sqrt{x})\) are calculated, you must simplify to aid further evaluation. Once the expression is reduced to manageable terms, substitute the limit value again to find the solution.

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

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. Evaluate this limit for the given value of \(x\) and function \(f.\) $$f(x)=3 x-4, \quad x=2$$

To prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. \(-\cos x=x \quad\) (one root). Make sure you are using radian mode.

Manufacturing electrical resistors Ohm's law for electrical circuits like the one shown in the accompanying figure states that \(V=R I .\) In this equation, \(V\) is a constant voltage, \(I\) is the current in amperes, and \(R\) is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which \(V\) will be 120 volts and \(I\) is to be \(5 \pm 0.1\) amp. In what interval does \(R\) have to lie for \(I\) to be within 0.1 amp of the value \(I_{0}=5 ?\) GRAPH CANT COPY Showing \(L\) is not a limit We can prove that \(\lim _{x \rightarrow c} f(x) \neq L\) by providing an \(\epsilon>0\) such that no possible \(\delta>0\) satisfies the condition $$\text { for all } x, \quad 0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon $$We accomplish this for our candidate \(\epsilon\) by showing that for each \(\delta>0\) there exists a value of \(x\) such that$$0<|x-c|<\delta \quad \text { and } \quad|f(x)-L| \geq \epsilon$$ GRAPH CANT COPY

Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$f(x)=x^{2}-5, \quad L=11, \quad c=4, \quad \epsilon=1$$

Explain why the equation \(\cos x=x\) has at least one solution.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.