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If \(2 x \leqslant g(x) \leqslant x^{4}-x^{2}+2\) for all \(x,\) evaluate \(\lim _{x \rightarrow 1} g(x)\)

Short Answer

Expert verified
The limit of \(g(x)\) as \(x\) approaches 1 is 2.

Step by step solution

01

Express the Inequality Constraints

The problem states that for all \(x\), \(2x \leq g(x) \leq x^4 - x^2 + 2\). We want to evaluate the limit of \(g(x)\) as \(x\) approaches \(1\).
02

Apply the Limit to Each Part of the Inequality

To find \(\lim_{{x \to 1}} g(x)\), we consider the limits of the boundary functions: \(2x\) and \(x^4 - x^2 + 2\) as \(x\) approaches \(1\).
03

Calculate the Left Boundary Limit

Calculate \(\lim_{{x \to 1}} 2x\): Since \(2x\) is continuous, \(\lim_{{x \to 1}} 2x = 2 \cdot 1 = 2\).
04

Calculate the Right Boundary Limit

For the right boundary, compute \(\lim_{{x \to 1}} (x^4 - x^2 + 2)\). Substituting \(x = 1\), we get \(1^4 - 1^2 + 2 = 2\).
05

Apply the Squeeze Theorem

Since \(2x \leq g(x) \leq x^4 - x^2 + 2\) for all \(x\), and both bounds approach 2 as \(x\) approaches 1, by the Squeeze Theorem, \(\lim_{{x \to 1}} g(x) = 2\).

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Key Concepts

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

Squeeze Theorem
The Squeeze Theorem is a valuable tool in calculus for finding the limit of a function when it is difficult to compute directly. It is particularly useful when a function is "trapped" between two other functions whose limits are known or easier to determine.
The theorem works by "squeezing" the function of interest between two simpler functions. If the upper and lower functions converge to the same limit at a certain point, then the function squeezed between them must also converge to that same limit.
  • First, you identify two functions, say \( f(x) \) and \( h(x) \), such that \( f(x) \leq g(x) \leq h(x) \) over an interval.
  • Next, compute \( \lim_{{x \to a}} f(x) \) and \( \lim_{{x \to a}} h(x) \).
  • If both limits are equal to a number \( L \), then \( \lim_{{x \to a}} g(x) = L \).
To apply the Squeeze Theorem successfully, make sure the inequality \( f(x) \leq g(x) \leq h(x) \) holds for values of \( x \) close to \( a \), except possibly at \( a \) itself.
Inequalities in Calculus
Understanding and working with inequalities is essential in calculus. Inequalities allow us to estimate and bound the behavior of functions, providing crucial insights into their properties.
In the context of limits, inequalities can help us establish upper and lower bounds for functions. This is especially useful when applying the Squeeze Theorem or similar techniques.
  • Inequalities can inform us about the continuity and boundedness of a function.
  • We often express functions in terms of inequalities to simplify the evaluation of a limit.
  • Understanding the interplay between functions using inequalities helps in establishing crucial relationships, such as dominance or convergence, as \( x \) approaches a particular value.
When dealing with inequalities, always check whether they remain valid over the interval of interest. These tools serve to create a bridge between complex expressions and simpler evaluations, leading to a greater understanding of the function's behavior.
Continuity of Functions
Continuity is a fundamental property in calculus that implies a function behaves predictably and smoothly without any jumps or interruptions.
A function is continuous at a point \( x = a \) if three conditions are satisfied:
  • The function \( f(x) \) is defined at \( a \).
  • The limit \( \lim_{{x \to a}} f(x) \) exists.
  • The limit and the function value are equal, i.e., \( \lim_{{x \to a}} f(x) = f(a) \).
Understanding and identifying continuity in functions is essential as it simplifies the process of finding limits. Continuous functions at a point or over an interval allow for straightforward limit calculations, where the limit can be evaluated by simple substitution.
Additionally, continuous functions are more predictable, which aids in applying theorems like the Intermediate Value Theorem, further emphasizing the importance of continuity in mathematical analysis and problem-solving.

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Most popular questions from this chapter

(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 \((\mathrm{c}) .\)

(a) If \(\mathrm{f}(\mathrm{t})=\mathrm{t}^{2}-\sqrt{\mathrm{t}},\) find \(\mathrm{f}^{\prime}(\mathrm{t}) .\) (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of \(\mathrm{f}\) and \(\mathrm{f}'.\)

Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically. $$\lim _{x \rightarrow 0} \frac{\sqrt{x+4}-2}{x}$$

$$ \begin{array}{c}{\text { (a) How large do we have to take } x \text { so that } 1 / x^{2}<0.0001 ?} \\ {\text { (b) Taking } r=2 \text { in Theorem } 5, \text { we have the statement }} \\ {\quad \lim _{x \rightarrow \infty} \frac{1}{x^{2}}=0} \\ {\text { Prove this directly using Definition } 7}\end{array} $$

(a) \(A\) tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the tank at a rate of 25 \(\mathrm{L} / \mathrm{min}\) . Show that the concentration of salt after t minutes (in grams per liter) is $$ \begin{array}{c}{\mathrm{C}(\mathrm{t})=\frac{30 \mathrm{t}}{200+\mathrm{t}}} \\ {\text { (b) What happens to the concentration as } \mathrm{t} \rightarrow \infty ?}\end{array} $$

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