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Chapter 11: Problem 51

What is wrong with the following use of I'H么pital's Rule: $$\lim _{x \rightarrow 1} \frac{x^{3}+x-2}{x^{2}-3 x+2}=\lim _{x \rightarrow 1} \frac{3 x^{2}+1}{2 x-3}=\lim _{x \rightarrow 1} \frac{6 x}{2}=3.$$ (The limit is actually \(-4 .\))

Short Answer

Expert verified
The given solution incorrectly applies L'H么pital's rule a second time. The correct limit after one application of the rule is \(-4\).

Step by step solution

01

Analyze the Original Limit

Consider the limit \ \ \[\lim_{x \to 1} \frac{x^3 + x - 2}{x^2 - 3x + 2}\]. \ Evaluate the numerator and denominator at \(x = 1\): \ \(1^3 + 1 - 2 = 0\) and \(1^2 - 3(1) + 2 = 0\). The limit is of the form \(\frac{0}{0}\), so we can apply L'H么pital's rule.
02

Apply L'H么pital's Rule Correctly

Differentiate the numerator and denominator separately: \ \ \[\text{Numerator: } \frac{d}{dx}(x^3 + x - 2) = 3x^2 + 1\] \ \[\text{Denominator: } \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3\]. \ Apply L'H么pital's rule: \ \[\lim_{x \to 1} \frac{x^3 + x - 2}{x^2 - 3x + 2} = \lim_{x \to 1} \frac{3x^2 + 1}{2x - 3}\].
03

Evaluate the New Limit

Evaluate the new limit by plugging \(x = 1\) into \(\frac{3x^2 + 1}{2x - 3}\): \ \ \[3(1)^2 + 1 = 4\] and \[2(1) - 3 = -1\]. \ Thus, \[\lim_{x \to 1} \frac{3x^2 + 1}{2x - 3} = \frac{4}{-1} = -4\].
04

Identify the Incorrect Step in the Solution Given

The given solution incorrectly applies L'H么pital's rule a second time, which is unnecessary and incorrect. After applying L'H么pital's rule once correctly, the limit \(\lim_{x \to 1} \frac{3x^2 + 1}{2x - 3}\) should be evaluated directly without further differentiation.

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Key Concepts

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

limit of a function
A limit of a function refers to the value that a function approaches as the input approaches a given point. To find limits, we analyze how the function behaves around the point of interest. For example, consider the limit \ \[ \lim_{x \to 1} \frac{x^3 + x - 2}{x^2 - 3x + 2} \]. When \( x = 1 \), both the numerator (\( x^3 + x - 2 \)) and the denominator (\( x^2 - 3x + 2 \)) become \( 0 \), creating the indeterminate form \( \frac{0}{0} \). Limits help us understand the behavior of functions when exact values aren't straightforward to calculate.
differentiation
Differentiation is a process in calculus that finds the rate at which a function is changing at any given point. It's essentially calculating the slope of the function's curve. In our exercise, to apply L'H么pital's Rule, we need to differentiate the numerator and the denominator separately: \( \frac{d}{dx}(x^3 + x - 2) = 3x^2 + 1 \) and \( \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3 \). Differentiating transforms our original limit into a form that is easier to evaluate. With these derivatives, we can re-evaluate the limit without the indeterminate form.
indeterminate forms
Indeterminate forms occur in calculus when substitution in a limit expression results in an undefined form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms mean that we can't directly calculate the limit and we need additional methods like L'H么pital's Rule. In our example, after substituting \( x = 1 \) into \( \frac{x^3 + x - 2}{x^2 - 3x + 2} \), we get the indeterminate form \( \frac{0}{0} \). Recognizing these forms is crucial because they tell us when to use techniques, like differentiation, to simplify and solve the limit. Using L'H么pital's Rule correctly transforms the indeterminate form into something manageable that does not need further differentiation.

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Most popular questions from this chapter

Suppose that \(f(0)=0\) and \(f^{\prime}\) is increasing. Prove that the function \(g(x)=\) \(f(x) / x\) is increasing on \((0 . \infty) .\) Hint: Obviously you should look at \(g^{\prime}(x)\) Prove that it is positive by applying the Mean Value Theorem to \(f\) on the right interval (it will help to remember that the hypothesis \(f(0)=0\) is essential, as shown by the function \(f(x)=1+x^{2}\) ).

(a) Suppose that the critical points of the polynomial function \(f(x)=x^{n}+\) \(a_{n-1} x^{n-1}+\dots+a_{0}\) are \(-1,1,2,3,\) and \(f^{\prime \prime}(-1)=0, f^{\prime \prime}(1)>0, f^{\prime \prime}(2)< 0, f^{\prime \prime}(3)=0 .\) Sketch the graph of \(f\) as accurately as possible on the basis of this information. (b) Does there exist a polynomial function with the above properties, except that 3 is not a critical point?

There is another form of I'H么pital's Rule which requires more than algebraic manipulations: If \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} g(x)=\infty,\) and \(\lim _{x \rightarrow \infty} f^{\prime}(x) / g^{\prime}(x)=l,\) then \(\lim _{x \rightarrow \infty} f(x) / g(x)=1 .\) Prove this as follows. (a) For every \(\varepsilon>0\) there is a number \(a\) such that $$\left|\frac{f^{\prime}(x)}{g^{\prime}(x)}-l\right|<\varepsilon \quad \text { for } x>a.$$ (Why can we assunc \(g(x)-g(a) \neq 0\) ? (b) Now write $$\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)} \cdot \frac{f(x)}{f(x)-f(a)} \cdot \frac{g(x)-g(a)}{g(x)}$$ (why can we assume that \(f(x)-f(a) \neq 0\) for large \(x\) ?) and conclude that \(\left|\frac{f(x)}{g(x)}-1\right|<2 \varepsilon \) for sulficiently large \(x.\)

Use derivatives to prove that if \(n \geq 1\), then $$(1+x)^{n}>1+n x \quad \text { for }-1< x < 0 \text { and } 0 < x$$ (notice that equality holds for \(x=0\) ).

(a) A point \(x\) is called a strict maximum point for \(f\) on \(A\) if \(f(x)>f(y)\) for all \(y\) in \(A\) with \(y \neq x\) (compare with the definition of an ordinary maximum point). A local strict maximum point is defined in the obvious way. Find all local strict maximum points of the function $$f(x)=\left\\{\begin{array}{ll} 0, & x \text { irrational } \\ \frac{1}{q}, & x=\frac{p}{q} \text { in lowest terms }. \end{array}\right.$$ It seems quite unlikely that a function can have a local strict maximum at every point (although the above example might give one pause for thought). Prove this as follows. (b) Suppose that every point is a local strict maximum point for \(f\). Let \(x_{1}\) be any number and choose \(a_{1} < x_{1} < b_{1}\) with \(b_{1}-a_{1}<1\) such that \(f\left(x_{1}\right)>f(x)\) for all \(x\) in \(\left[a_{1}, b_{1}\right] .\) Let \(x_{2} \neq x_{1}\) be any point in \(\left(a_{1}, b_{1}\right)\) and choose \(a_{1} \leq a_{2}< x_{2} < b_{2} \leq b_{1}\) with \(b_{2}-a_{2} < \frac{1}{2}\) such that \(f\left(x_{2}\right)>f(x)\) for all \(x\) in \(\left[a_{2}, b_{2}\right] .\) Continue in this way, and use the Nested Interval Theorem (Problem 8-14 ) to obtain a contradiction.
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