/*! 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 22 $$ \text { Find } \lim _{h \ri... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 22

$$ \text { Find } \lim _{h \rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h} $$

Short Answer

Expert verified
The limit as \(h\) approaches \(0\) in \(\frac{1/x+h - 1/x}{h}\) is \(-\frac{1}{x^2}\)

Step by step solution

01

Simplify Denominator

The first step is to simplify the expression. To do this, remove \(h\) from the denominator by getting a common denominator for the fractions in the numerator. Simplify \(\frac{1}{x+h}-\frac{1}{x}\) by finding a common denominator: \[ \frac{1}{x+h}-\frac{1}{x} = \frac{x-(x+h)}{x(x+h)} = -\frac{h}{x(x+h)} \] The expression simplifies to: \[ \frac{-h/(x(x+h))}{h} \]
02

Cancel Terms

Now, cancel out the \(h\) in the numerator and denominator to get: \[ -\frac{1}{x(x+h)} \] Now the expression is a simple fraction and we can apply the limit.
03

Apply the Limit

Finally, take the limit as \(h\) approaches 0. By substituting \(h = 0\) into the final expression from step 2, find the limit: \[ \lim_{h \rightarrow 0} -\frac{1}{x(x+h)} = -\frac{1}{x^2} \]This result is negative because of the subtraction in the original expression.

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.

Understanding Continuity in Calculus
Continuity is a fundamental concept in calculus that describes how smooth a function is at a point. If a function is continuous at a point, it means there are no breaks, jumps, or holes at that location.
In simple terms, a continuous function can be drawn without lifting your pencil from the paper. This idea is crucial as it ensures that limits behave nicely.

In our exercise, we're working with the function \( f(x) = \frac{1}{x} \). At any point where the function is defined, particularly away from zero, the function should exhibit continuity. This allows us to calculate limits by directly substituting values, once we've simplified the expression.
  • If a function is not continuous, limits can become more complex, often requiring additional methods or approaches.
  • Discontinuities often occur with division by zero, which is why \( x eq 0 \) is essential for our exercise.
Rational Expressions and Their Simplification
Rational expressions involve ratios of polynomials, like \( \frac{1}{x} \) or \( \frac{1}{x+h} \). Understanding how to handle these expressions is key, especially in calculus where limits and derivatives often involve them.
When dealing with rational expressions, finding a common denominator is a useful technique. It allows us to combine or simplify fractions easily.
  • In our problem, the expression \( \frac{1}{x+h} - \frac{1}{x} \) is simplified by using a common denominator: \( x(x+h) \).
  • Simplifying leads to cleaner expressions, clearing away complexity for more straightforward limit evaluation.
Eventually, we cancel terms, which is a common step in handling rational expressions, to simplify further before applying limits or derivatives.
Introduction to Derivatives
Derivatives represent the rate of change of a function. They are a cornerstone of calculus and provide insight into function behaviors, optimization, and more.

The derivative of a function at a point gives the slope of the tangent line to the graph at that point.
In our exercise, finding \( \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \) is actually the definition of the derivative, applied to \( f(x) = \frac{1}{x} \).
  • This limit evaluates the instantaneous rate of change of the function, which is the derivative.
  • In our example, by simplifying and evaluating the limit, we find that this rate of change is \( -\frac{1}{x^2} \).
This negative sign indicates that as \( x \) increases, the value of the function decreases, reflecting the inverse relationship emphasized in rational expressions.

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

\(\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{6}+h\right)-\tan \left(\frac{\pi}{6}\right)}{h}=\) (A) \(\frac{4}{3}\) (B) \(\sqrt{3}\) (C) 0 (D) \(\frac{3}{4}\)

Evaluate \(\int_{0}^{1}\left(x^{4}-5 x^{3}+3 x^{2}-4 x-6\right) d x\)

\(\int x \sqrt{5 x^{2}-4} d x=\) (A) \(\frac{1}{10}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C\) (B) \(\frac{1}{15}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C\) (C) \(\frac{20}{3}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C\) (D) \(\frac{3}{20}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C\)

Water is draining at the rate of 48\(\pi \mathrm{f}^{3} / \mathrm{second}\) from the vertex at the bottom of a conical tank whose diameter at its base is 40 feet and whose height is 60 feet. (a) Find an expression for the volume of water in the tank, in terms of its radius, at the surface of the water. (b) At what rate is the radius of the water in the tank shrinking when the radius is 16 feet? (c) How fast is the height of the water in the tank dropping at the instant that the radius is 16 feet?

Now evaluate the following integrals. \(\int \sin (\sin x) \cos x d x\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision 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.