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Chapter 1: Problem 6

$$\text { Show that } \lim _{t \rightarrow 1} \frac{t^{3}-t^{2}-5 t}{6-t^{2}}=-1$$

Short Answer

Expert verified
The limit is \(-1\) by direct substitution.

Step by step solution

01

Identify the Problem and Limit Formula

We need to evaluate the limit \( \lim_{{t \to 1}} \frac{{t^3 - t^2 - 5t}}{{6 - t^2}} \). Start by trying to substitute \( t = 1 \) directly into the expression.
02

Substitute and Check Direct Substitution

If we substitute \( t = 1 \) directly, we get the expression \( \frac{{1^3 - 1^2 - 5(1)}}{{6 - 1^2}} = \frac{{1 - 1 - 5}}{{6 - 1}} = \frac{-5}{5} = -1 \). Thus, direct substitution shows that the limit of the expression as \( t \to 1 \) is \(-1\).
03

Confirm by Factoring and Simplifying (Verification)

Factor the numerator, \( t^3 - t^2 - 5t \), to verify if direct substitution could have involved cancellation. The expression becomes \( t(t^2 - t - 5) \). Since there are no common factors with the denominator \( 6 - t^2 \), direct substitution already provides a straightforward answer.

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Key Concepts

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

Direct substitution
Direct substitution is often the first method used when evaluating limits. It involves simply substituting the variable with the value it approaches, in this case, substituting \( t = 1 \) directly into the expression \( \frac{t^3 - t^2 - 5t}{6 - t^2} \). The key here is to check if this substitution leads to a determinate form, such as a non-zero over non-zero result.
  • First, substitute the value: \( \frac{1^3 - 1^2 - 5 \times 1}{6 - 1^2} = \frac{-5}{5} = -1 \).
  • If the expression results in an indeterminate form like \( \frac{0}{0} \), further techniques are needed. But here, direct substitution conveniently provides the limit of \(-1\).
Direct substitution is the fastest method when applicable. If it returns a valid number without simplification, any additional steps are unnecessary.
Numerator factoring
Factoring is a crucial algebraic tool that helps in simplifying expressions, especially when dealing with polynomial limits. For our given problem, factoring the numerator could reveal hidden simplifications.Numerator factoring involves breaking down the polynomial \( t^3 - t^2 - 5t \) into factorable components. We factored it as \( t(t^2 - t - 5) \).
  • Start by identifying common factors. Here, \( t \) is common, resulting in \( t(t^2 - t - 5) \).
  • Check the numerator and denominator for common factors that may simplify the fraction.
In our exercise, factored or simplified terms did not cancel with the denominator. This confirmed that direct substitution was sufficient and correct. Factoring is a powerful method to ensure there are no overlooked cancellations.
Limit verification
Limit verification is the process of confirming that the evaluated limit is correct, often through alternative methods. In the provided exercise, we primarily used direct substitution, but another form of verification can be beneficial.
  • After direct substitution, verify the result is consistent through factoring or other methods such as conjugate multiplication or rationalization.
  • Especially ensure no common factors in the numerator and denominator are missed which might initially cause a wrong indeterminate form.
By validating the direct substitution process with factoring, we verified that the limit \(-1\) is indeed correct. Verification reassures the results by cross-checking with different techniques, highlighting the robustness of the solution done initially.

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Most popular questions from this chapter

Calculate the slope of the line through each pair of points. a. (2,7),(-3,-8) b. \(\left(\frac{1}{2}, \frac{3}{2}\right),\left(\frac{7}{2},-\frac{7}{2}\right)\) c. (6.3,-2.6),(1.5,-1)

A manufacturer of soccer balls finds that the profit from the sale of \(x\) balls per week is given by \(P(x)=160 x-x^{2}\) dollars. a. Find the profit on the sale of 40 soccer balls per week. b. Find the rate of change in profit at the production level of 40 balls per week. c. Using a graphing calculator, graph the profit function and, from the graph, determine for what sales levels of \(x\) the rate of change in profit is positive.

Sketch a graph of the following function: \(f(x)=\left\\{\begin{array}{c}x^{2}, \text { if } x<0 \\ 3, \text { if } x \geq 0\end{array}\right.\) Is the function continuous?

Consider the function \(f(x)=x^{3}\). a. Copy and complete the following table of values. \(P\) and \(Q\) are points on the graph of \(f(x)\). $$\begin{array}{|c|c|c|} \hline \boldsymbol{p} & \boldsymbol{Q} & \begin{array}{l} \text { Slope of line } \\ \text { PQ } \end{array} \\ \hline(2,) & (3,) & \\ \hline(2,) & (2.5,) & \\ \hline(2,) & (2.1,) & \\ \hline(2,) & (2.01,) & \\ \hline(2,) & (1,) & \\ \hline(2,) & (1.5,) & \\ \hline(2,) & (1.9,) & \\ \hline(2,) & (1.99,) & \\ \hline \end{array}$$ b. Use your results for part a to approximate the slope of the tangent to the graph of \(f(x)\) at point \(P\). c. Calculate the slope of the secant \(P Q\), where the \(x\) -coordinate of \(Q\) is \(2+h\). d. Use your result for part \(c\) to calculate the slope of the tangent to the graph of \(f(x)\) at point \(P\). e. Compare your answers for parts b and d. f. Sketch the graph of \(f(x)\) and the tangent to the graph at point \(P\).

A spaceship approaching touchdown on a distant planet has height \(h,\) in metres, at time \(t,\) in seconds, given by \(h=25 t^{2}-100 t+100 .\) When does the spaceship land on the surface? With what speed does it land (assuming it descends vertically)?
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