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Chapter 2: Problem 57

Find the limits in Exercises \(53-58 .\) \begin{equation} \lim \frac{x^{2}-3 x+2}{x^{3}-2 x^{2}} \end{equation} \begin{equation} \begin{array}{ll}{\text { a. }} & {x \rightarrow 0^{+}} \quad {\text { b. } x \rightarrow 2^{+}} \\ {\text { c. }} & {x \rightarrow 2^{-}} \quad {\text { d. } x \rightarrow 2} \\ {\text { e. }} & {\text { What, if anything, can be said about the limit as } x \rightarrow 0 ?}\end{array} \end{equation}

Short Answer

Expert verified
a. \( -\infty \); b. \( \infty \); c. \( -\infty \); d. The limit does not exist; e. \( -\infty \).

Step by step solution

01

Simplify the given expression

We start with the expression \( \frac{x^2 - 3x + 2}{x^3 - 2x^2} \). Factorize both the numerator and the denominator. The numerator factors to \( (x - 1)(x - 2) \) and the denominator factors to \( x^2(x - 2) \). Therefore, the expression simplifies to \( \frac{(x-1)}{x^2} \) as long as \( x eq 2 \).
02

Find the limit as \( x \rightarrow 0^{+} \)

Now that the expression is \( \frac{(x-1)}{x^2} \), calculate the limit as \( x \rightarrow 0^+ \). As \( x \) approaches zero from the positive side, \( (x-1) \) approaches \(-1\), and \( x^2 \) approaches \(0^+\). This leads the limit to approach \( -\infty \). Therefore, \( \lim_{x \to 0^+} \frac{(x-1)}{x^2} = -\infty \).
03

Determine the limit as \( x \rightarrow 2^{+} \)

With the simplified expression \( \frac{(x-1)}{x^2} \), calculate the limit as \( x \) approaches 2 from the positive side. While the simplification worked for expression, consider original division. Substitute back \( x \) close to 2 like \( x^2-2x \) leads the denominator of \( \frac{x-1}{x(x-2)} \) going to zero from positive. Result is \( +\infty \). Thus, \( \lim_{x \to 2^+} \frac{x^2 - 3x + 2}{x^3 - 2x^2} = +\infty \).
04

Determine the limit as \( x \rightarrow 2^{-} \)

Consider limit for \( \frac{x-1}{x^2} \) keeping role of real expression affected near forbidden area. Investigate not by factor simplification using \( x \rightarrow 2^- \). The term \( x(x-2) \) leads denominator to zero, \( x \rightarrow 2^- \) implies approach \(-\infty\), Therefore, \( \lim_{{x \to 2^-}} \frac{x^2 - 3x + 2}{x^3 - 2x^2} = -\infty \).
05

Evaluate the limit as \( x \rightarrow 2 \)

A discontinuity exists at \( x = 2 \), as from above parts, differing approaches as split directions into positive \((+\infty)\) and negative \((-\infty)\), making the overall limit as doesn鈥檛 exist. Thus, \( \lim_{x \to 2} \frac{x^2 - 3x + 2}{x^3 - 2x^2} \) does not exist.
06

Analyze the limit as \( x \rightarrow 0 \) (encompass cases around zero)

The original context over means checking general zero movement, still leading to changes due mostly smaller scales upper realize positive \( \lim_{{x \rightarrow 0}} x/x^2\). Having adjusted, ensure understanding around differing treatment concluding potentially but maintaining focus historical provided limits leading loss continual balance resulting. Thus \( \lim_{{x \to 0}} = -\infty \).

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Key Concepts

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

Rational Functions
Rational functions are expressions that can be written as a ratio of two polynomials, such as \( \frac{f(x)}{g(x)} \). These expressions are useful in calculus to analyze behavior and trends of complex equations. They allow us to identify key features, such as asymptotes and intercepts, by examining their numerators and denominators individually.
  • Numerator: determines zeros or roots of the function where the output is zero.
  • Denominator: indicates potential asymptotic behavior, especially since division by zero is undefined.
Simplifying rational functions by factorizing is crucial for finding these key features. It can enable us to see the function's behavior near points of interest, such as where the denominator might be zero, though simplifications can't be used if the denominator is exactly zero.
Factorization in Calculus
Factorization is breaking down expressions into simpler multipliers that are easier to solve or analyze. In calculus, factorizing rational functions helps simplify expressions and find limits:
  • The expression \( \frac{x^2 - 3x + 2}{x^3 - 2x^2} \) can be factorized to \( \frac{(x-1)(x-2)}{x^2(x-2)} \).
  • Cancellation possible only if the common factor isn鈥檛 zero. We get \( \frac{x-1}{x^2} \) when \( x eq 2 \).
This factorizes both numerator and denominator, revealing asymptotes or discontinuities in the function. Factorization is especially helpful for determining limits, where simplifying expressions can help us avoid the pitfalls of zero in the denominator.
One-Sided Limits
One-sided limits involve approaching a particular point from only one direction: either from the left \((x^-_+)\) or the right \((x^+_-)\). They are essential in understanding specific behavior at points of interest in rational functions.
For example, evaluating the limit of the function as \( x \to 0^+ \), we see:
  • Numerator tends towards \(-1\)
  • Denominator tends towards a small positive value, creating a result tending towards \(-\infty\)
One-sided limits provide insight into whether discontinuities exist as sharp jumps or if the function diverges in certain directions. They allow us to pinpoint exactly where discrepancies and continuous functions meet.
Continuity and Discontinuity
Continuity in calculus means a function behaves predictably without abrupt changes, while discontinuity often signifies points of division or jumps. Rational functions commonly have discontinuities, especially when their denominators equal zero.
  • At \( x = 2 \) in the function \( \frac{x-1}{x^2} \), there鈥檚 discontinuity because the limit from each side differs: \(+\infty\) from the right and \(-\infty\) from the left.
  • When the left and right limits agree, the function is considered continuous at that point.
Understanding whether a function is continuous or discontinuous at critical points helps in assessing the overall behavior and defining limits accurately.
Infinite Limits
An infinite limit describes the behavior of a function as it grows larger and larger in magnitude, typically occurring when the denominator approaches zero, causing the function to diverge. In the case of \( \lim_{{x \to 0^+}} \frac{(x-1)}{x^2} \), the denominator decreases rapidly towards zero, making the entire function grow negatively without bound, signified by \(-\infty\). As such,
  • Positive or negative infinite limits reveal asymptotic tends in functions.
  • Data around these infinity points can help us outline the general flow of graphs and rational functions.
Recognizing these infinite tendencies is key in analyzing behavior at boundary lines where functions might not physically cross but move infinitely close toward.

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Most popular questions from this chapter

Find the limits in Exercises \(84-90 .\) (Hint: Try multiplying and dividing by the conjugate.) $$ \lim _{x \rightarrow \infty}(\sqrt{x+9}-\sqrt{x+4}) $$

Graph the rational functions in Exercises \(63-68 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{1}{x+1} $$

Find the limits. \begin{equation}\lim _{\theta \rightarrow 0} \theta \cos \theta\end{equation}

Use the definitions of right-hand and left-hand limits to prove the limit statements. \begin{equation}\lim _{x \rightarrow 2^{+}} \frac{x-2}{|x-2|}=1\end{equation}

Suppose that \(f\) is an even function of \(x .\) Does knowing that \(\lim _{x \rightarrow 2^{-}} f(x)=7\) tell you anything about either \(\lim _{x \rightarrow-2^{-}} f(x)\) or \(\lim _{x \rightarrow-2^{+}} f(x) ?\) Give reasons for your answer.
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