/*! 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 50 Evaluating a Limit from Calculus... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluating a Limit from Calculus $$ \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $$ $$ f(x)=\frac{1}{x-1} $$

Short Answer

Expert verified
\(-\frac{1}{(x-1)^2}\

Step by step solution

01

Substitute Function

Replace \(f(x+h)\) and \(f(x)\) in the limit equation with our function value. So, we get \(\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)-1}-\frac{1}{x-1}}{h}\)
02

Simplify the Numerator

The numerator can be simplified by finding a common denominator. The common denominator is \((x+h-1)(x-1)\), multiply top and bottom of each fraction by this denominator to remove the fractions within a fraction, \(\lim _{h \rightarrow 0} \frac{(x-1)-(x+h-1)}{h(x-1)(x+h-1)}\)
03

Simplify the Expression

Simplify the expression \(\lim _{h \rightarrow 0} \frac{-h}{h(x+h-1)(x-1)}\), the -h in the numerator can cancel out the h in the denominator, we are left with \(\lim _{(h \rightarrow 0)} \frac{-1}{(x+h-1)(x-1)}\)
04

Apply the Limit

Finally we apply the limit, by putting h=0 in the equation, we obtain the limit: \(\frac{-1}{(x-1)^2}\)

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Key Concepts

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

Limits in Calculus
Understanding limits is essential in calculus, as they are the foundation for the study of continuity, derivatives, and integrals. In simple terms, a limit describes the value that a function approaches as the input (or argument) of the function approaches some value.

For instance, in the exercise \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\), we are interested in finding the value that the expression approaches as \(h\) becomes very close to zero. This type of limit, known as the difference quotient, is specifically important because, when it exists, it defines the derivative of \(f(x)\) at \(x\). It measures how the function value changes as \(h\), which represents a tiny change in \(x\), approaches zero. Calculating such limits accurately is critical for analyzing and understanding the behavior of functions in calculus.
Limit of a Function
The limit of a function is a fundamental concept in calculus that describes the behavior of a function as its argument approaches a particular value. For example, in the provided exercise, the goal is to evaluate \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\), which is essentially asking for the behavior of the fraction as \(h\) gets infinitely close to zero.

If a function approaches a specific value \(L\) as \(x\) approaches \(c\) from either side, the limit of the function as \(x\) approaches \(c\) is \(L\). We write this as \(\lim _{x \rightarrow c} f(x) = L\). By mastering the evaluation of function limits, students gain insight into the continuity of functions and the beginnings of derivative concepts.
Simplifying Expressions
Simplifying expressions in calculus often involves techniques from algebra that help transform complex expressions into simpler ones. The numerator or denominator in fractions, for instance, can be simplified by combining like terms or using common denominators.

In our exercise step related to \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\), an essential simplification step involves finding a common denominator for the fractions within the limit expression. After obtaining the common denominator, the next step is to cancel out terms. When \(h\) appears both in the numerator and the denominator, and the limit approaches zero, we can often cancel out the \(h\) term safely, further simplifying the expression and making it easier to evaluate the limit. These algebraic skills are extremely important for solving more complex calculus problems efficiently.
Precalculus Mathematics
Precalculus encompasses various mathematical principles that prepare students for the study of calculus. These principles include algebraic manipulation, functions, complex numbers, and the concept of limits. A solid understanding of precalculus is critical for success in calculus because it lays the groundwork for all the major topics within it.

For example, the limit problem given in the exercise requires precalculus knowledge such as finding a common denominator for fractions and simplifying complex algebraic expressions. These skills are essential for effectively managing the more challenging aspects of limits and differentiation in calculus. Focusing on precalculus can ease the transition for students moving into calculus, as they will already be familiar with the necessary foundational tools.

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Most popular questions from this chapter

Finding the Limit of a Sequence In Exercises \(45 - 54\) , write the first five terms of the sequence and find the limit of the sequence (if it exists). If the limit does not exist, then explain why. Assume \(n\) begins with 1 . $$ a _ { n } = \frac { n } { 2 n + 1 } $$

Determine whether the statement is true or false. Justify your answer. The slope of the graph of \(y=x^{2}\) is different at every point on the graph of \(f .\)

Social Security The average monthly Social Security benefit \(B\) (in dollars) for a retired worker age 62 or over from 2001 through 2010 is given by $$B = \frac { 855.0 - 116.74 t + 5.565 t ^ { 2 } } { 1.0 - 0.16 t + 0.008 t ^ { 2 } } , \quad 1 \leq t \leq 10$$ where \(t\) represents the year, with \(t = 1\) corresponding to \(2001 .\) (Source: U.S. Social Security Administration) (a) Use the model to predict the average monthly benefit in \(2017 .\) (b) What is the limit of the function as \(t\) approaches infinity? Explain the meaning of the limit in the context of the problem. (c) Discuss why this model is not realistic for long-term predictions of average monthly Social Security benefits.

Evaluating a Limit at Infinity In Exercises \(9 - 28\) , find the limit (if it exists). If the limit does not exist, then explain why. Use a graphing utility to verify your result graphically. $$ \lim _ { x \rightarrow \infty } \frac { 2 x ^ { 2 } - 6 } { ( x - 1 ) ^ { 2 } } $$

Determining Convergence or Divergence In Exercises \(65 - 68\) , create a scatter plot of the terms of the sequence. Determine whether the sequence converges or diverges. When it converges, estimate its limit. $$ a _ { n } = 3 \left( \frac { 3 } { 2 } \right) ^ { n } $$

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