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Chapter 2: Problem 12

\(10-12\) Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a. \(h(t)=\frac{2 t-3 t^{2}}{1+t^{3}}, \quad a=1\)

Short Answer

Expert verified
The function is continuous at \(t = 1\) because \(\lim_{t \to 1} h(t) = h(1) = -\frac{1}{2}\).

Step by step solution

01

Understanding Continuity

A function is continuous at a point \(a\) if \(\lim_{t \to a} h(t) = h(a)\). We need to calculate both \(h(1)\) and \(\lim_{t \to 1} h(t)\), and show that they are equal.
02

Computing \(h(1)\)

Substitute \(t = 1\) into the function \(h(t)\): \[h(1) = \frac{2(1) - 3(1)^2}{1 + (1)^3} = \frac{2 - 3}{1 + 1} = \frac{-1}{2}\] So, \(h(1) = -\frac{1}{2}\).
03

Calculating the Limit \(\lim_{t \to 1} h(t)\)

Compute the limit: \[\lim_{t \to 1} \frac{2t - 3t^2}{1 + t^3}\] To do this, merge terms and simplify if needed. The function itself is a rational function and, at \(t = 1\), the expression does not become undefined, so:\[ = \frac{2(1) - 3(1)^2}{1 + 1^3} = \frac{-1}{2}\].
04

Comparing Limit and Function Value

Now, compare \(\lim_{t \to 1} h(t)\) and \(h(1)\):\[\lim_{t \to 1} h(t) = -\frac{1}{2} = h(1)\].Both are equal, confirming the function is continuous at \(t = 1\).

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Key Concepts

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

Definition of Continuity
Continuity is a fundamental concept in calculus that helps us understand the behavior of functions. To say that a function is continuous at a point means there are no breaks, jumps, or holes at that point in the graph of the function. Specifically, a function \( f(x) \) is continuous at a point \( a \) if three conditions are met:
  • The function \( f(x) \) is defined at \( a \).
  • The limit \( \lim_{x \to a} f(x) \) exists.
  • The limit equals the function's value at \( a \): \( \lim_{x \to a} f(x) = f(a) \).
In our specific case with \(h(t)\), we've confirmed it is continuous at \( t = 1 \) by finding both \( h(1) \) and \( \lim_{t \to 1} h(t) \), and checking they are equal.
Properties of Limits
Limits play a crucial role in determining continuity and overall function behavior. The limit describes what the function approaches as the input approaches some value. Several properties of limits help simplify the computation:
  • Addition/Subtraction: \( \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \)
  • Multiplication: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \)
  • Division: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \) if \( \lim_{x \to a} g(x) eq 0 \)
By applying these properties, we evaluated the limit \( \lim_{t \to 1} \frac{2t - 3t^2}{1 + t^3} \) without complications because the denominator is not zero at \( t = 1 \). This confirms the functioning of these limit properties when assessing continuity.
Rational Functions
Rational functions are quotients of polynomials, like \( h(t) = \frac{2t - 3t^2}{1 + t^3} \). These kinds of functions have unique characteristics due to being composed of polynomials:
  • They are continuous wherever the denominator is non-zero, which ensures the function is defined.
  • Rational functions can be simplified by cancelling common factors, given they don't turn the denominator into zero.
  • The limit of a rational function as \( t \) approaches a value usually involves direct substitution unless the function is undefined at the point.
In the example provided, since the denominator \(1 + t^3\) is not zero at \( t = 1 \), the function \( h(t) \) remains continuous there. Understanding these features makes evaluating and proving the continuity straightforward, as was shown in the solution by directly substituting \( t = 1 \) into the function.

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Most popular questions from this chapter

The point \(P(1,0)\) lies on the curve \(y=\sin (10 \pi / x)\). $$\begin{array}{l}{\text { (a) If } Q \text { is the point }(x, \sin (10 \pi / x)) \text { , find the slope of the secant }} \\ {\text { line } P Q \text { (correct to four decimal places) for } x=2,1.5,1.4 \text { , }} \\\ {1.3,1.2,1.1,0.5,0.6,0.7,0.8, \text { and } 0.9 . \text { Do the slopes }} \\\ {\text { appear to be approaching a limit? }}\end{array}$$ $$\begin{array}{l}{\text { (b) Use a graph of the curve to explain why the slopes of the }} \\ {\text { secant lines in part (a) are not close to the slope of the tan- }} \\ {\text { gent line at } P .}\end{array}$$ $$\begin{array}{l}{\text { (c) By choosing appropriate secant lines, estimate the slope of }} \\ {\text { the tangent line at } \mathrm{P} .}\end{array}$$

Graph the function \(f(x)=x+\sqrt{|x|}\) . Zoom in repeatedly, first toward the point \((-1,0)\) and then toward the origin. What is different about the behavior of \(f\) in the vicinity of these two points? What do you conclude about the differentiability of \(f\)?

(a) Evaluate h \((\mathrm{x})=(\tan \mathrm{x}-\mathrm{x}) / \mathrm{x}^{3}\) for \(\mathrm{x}=1,0.5,0.1,0.05\) \(0.01,\) and 0.005 (b) Guess the value of \(\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}\) (c) Evaluate h( \(x\) ) for successively smaller values of \(x\) until you finally reach a value of 0 for \(h(x)\) . Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained a value of \(0 .\) (In Section 4.4 a method for evaluating the limit will be explained.) (d) Graph the function \(h\) in the viewing rectangle \([-1,1]\) by \([0,1]\) . Then zoom in toward the point where the graph crosses the y-axis to estimate the limit of \(h(x)\) as \(x\) approaches \(0 .\) Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part \((\mathrm{c}) .\)

Make a careful sketch of the graph of \(f\) and below it sketch the graph of \(f^{\prime}\) in the same manner as in Exercises \(4-11\) . Can you guess a formula for \(f^{\prime}(x)\) from its graph? \(f(x)=e^{x}\)

Determine the infinite limit. $$\lim _{x \rightarrow \pi^{-}} \cot x$$
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