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Chapter 2: Problem 79

If \(\lim _{x \rightarrow 4} \frac{f(x)-5}{x-2}=1,\) find \(\lim _{x \rightarrow 4} f(x).\)

Short Answer

Expert verified
\( \lim _{x \rightarrow 4} f(x) = 5 \).

Step by step solution

01

Understand the Problem

We are given the limit \( \lim _{x \rightarrow 4} \frac{f(x)-5}{x-2}=1 \) and need to find \( \lim _{x \rightarrow 4} f(x) \). This problem involves a limit that results in L'H么pital's Rule since the expression is an indeterminate form \( \frac{0}{0} \).
02

Identify the Form of the Limit

The expression \( \frac{f(x)-5}{x-2} \) suggests that both the numerator \( f(x) - 5 \) and the denominator \( x - 2 \) approach zero as \( x \rightarrow 4 \). This implies \( f(x) - 5 \) must approach \( 0 \), meaning \( f(x) \) approaches some constant that needs to be determined.
03

Apply L'H么pital's Rule

Since \( \frac{f(x)-5}{x-2} \) is a \( \frac{0}{0} \) form, we apply L'H么pital's Rule. This states that the limit of \( \frac{f(x)-5}{x-2} \) is the same as the limit of \( \frac{f'(x)}{1} \) or \( f'(x) \) as \( x \to 4 \). However, L'H么pital's Rule is not needed since the limit condition given is enough.
04

Set Up the Equation for the Limit

From \( \lim _{x \rightarrow 4} \frac{f(x)-5}{x-2}=1 \), it suggests the derivative condition \( \frac{f(x)-5}{x-4} = 1 \) can be written as \( f(x)-5 = 1(x-4) \).
05

Solve for f(x)

Solve the equation \( f(x)-5 = 1(x-4) \) to find \( f(x) = x - 4 + 5 = x + 1 \).
06

Find the Required Limit of f(x)

Evaluate \( \lim _{x \rightarrow 4} f(x) \) by substituting into \( f(x) = x + 1 \): \( \lim _{x \rightarrow 4} (x+1) = 4+1 = 5 \).

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Key Concepts

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

L'H么pital's Rule
L'H么pital's Rule is a powerful tool in calculus used to evaluate limits that initially present as indeterminate forms. This rule applies to limits of the form \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). When faced with such a scenario:
  • Differentiate the numerator and the denominator separately.
  • Re-evaluate the limit using these derivatives.
If after differentiation, the resulting expression is still an indeterminate form, repeat the process until it can be evaluated. However, it's essential to check if L'H么pital's Rule is needed, as some problems can be solved directly by simplifying or algebraic manipulation, like in this case where the problem already provides enough information to find the limit directly, bypassing the need for derivatives altogether.
Indeterminate Forms
Indeterminate forms occur when substituting values into a function's limit results in expressions that are not immediately clear or determinate. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \), \( 0 \times \infty \), among others.These forms are tricky because they do not provide enough information to determine a limit immediately. However:
  • They signal the need for additional techniques like algebraic manipulation, L'H么pital's Rule, or approximation methods.
  • Analyzing the structure of the expressions involved helps decide which technique to use.
By recognizing these forms, you can more effectively employ mathematical strategies to solve the limit problem.
Limit Calculation
Calculating limits involves finding the value that a function approaches as the input (or variable) approaches a particular point. To find limits:
  • Sometimes direct substitution suffices if the function is continuous at the point.
  • For functions that result in an indeterminate form upon substitution, apply algebraic simplifications or rules like L'H么pital's to find the answer efficiently.
In our exercise, the problem was resolved directly by rewriting the equation \( f(x)-5 = 1(x-4) \) to \( f(x) = x + 1 \) and substituting \( x = 4 \). This straightforward approach shows how understanding the structure of limits can lead to simpler solutions without more complex calculus techniques.

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Most popular questions from this chapter

Suppose that \(g(x) \leq f(x) \leq h(x)\) for all \(x \neq 2\) and suppose that $$\lim _{x \rightarrow 2} g(x)=\lim _{x \rightarrow 2} h(x)=-5.$$ Can we conclude anything about the values of \(f, g,\) and \(h\) at \(x=2 ?\) Could \(f(2)=0 ?\) Could \(\lim _{x \rightarrow 2} f(x)=0 ?\) Give reasons for your answers.

Show that the equation \(x^{3}-15 x+1=0\) has three solutions in the interval [-4,4]

a. Graph \(g(x)=x \sin (1 / x)\) to estimate \(\lim _{x \rightarrow 0} g(x),\) zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof.

a. Graph \(h(x)=x^{2} \cos \left(1 / x^{3}\right)\) to estimate \(\lim _{x \rightarrow 0} h(x),\) zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof.

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
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