Domanda del libro di testo
Complete the following steps for each function.
c. State the interval(s) of continuity.
f(x)={x^3+4x+1 if x≤0
2x^3 if x>0; a=0
243
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Ch. 2 - Limits
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 41
Find the following limits or state that they do not exist. Assume a, b , c, and k are fixed real numbers.
lim x→9 √x − 3 / x − 9
Guida verificata passo dopo passo1
Recognize that the limit \( \lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} \) is an indeterminate form \( \frac{0}{0} \).
To resolve the indeterminate form, use algebraic manipulation. Multiply the numerator and the denominator by the conjugate of the numerator: \( \sqrt{x} + 3 \).
This gives: \( \lim_{x \to 9} \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} \).
Simplify the expression: the numerator becomes \( x - 9 \) because \((\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9\).
Cancel \( x - 9 \) from the numerator and denominator, resulting in \( \lim_{x \to 9} \frac{1}{\sqrt{x} + 3} \).
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Limits
A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It helps in understanding how functions behave near points of interest, including points where they may not be defined. In this case, we are interested in the limit of a function as x approaches 9.
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One-Sided Limits
Indeterminate Forms
Indeterminate forms occur in calculus when direct substitution in a limit leads to an ambiguous result, such as 0/0. In the given limit, substituting x = 9 results in this form, necessitating further analysis, such as algebraic manipulation or applying L'Hôpital's Rule to resolve the limit.
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3:56Slope-Intercept Form
Rationalizing the Numerator
Rationalizing the numerator is a technique used to simplify expressions involving square roots. By multiplying the numerator and denominator by the conjugate of the numerator, we can eliminate the square root and simplify the limit calculation. This method is particularly useful when dealing with limits that yield indeterminate forms.
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6:47Finding Limits Numerically and Graphically
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