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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 2.2.51

Using Limit Rules


Suppose lim x→0 f(x) = 1 and lim x→0 g(x) = −5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.


limx→0 (2f(x) − g(x)) / (f(x) + 7)² = limx→0 (2f(x) − g(x)) / limx→0 (f(x) + 7)² (a)


(We assume the denominator is nonzero.)


(lim x→0 2f(x) − lim x→0 g(x)) / (lim x→0 (f(x) + 7))² (b)


= (2 lim x→0 f(x) − lim x→0 g(x)) / (lim x→0 f(x) + lim x→0 7)² (c)


= ((2)(1) − (−5)) / (1 + 7)² = 7/64

Guida verificata passo dopo passo
1
Step 1: Identify the limit expression given: lim_{x→0} (2f(x) − g(x)) / (f(x) + 7)². We need to apply limit rules to simplify this expression.
Step 2: Apply the Quotient Rule for limits, which states that lim_{x→a} [u(x)/v(x)] = [lim_{x→a} u(x)] / [lim_{x→a} v(x)], provided lim_{x→a} v(x) ≠ 0. This allows us to separate the limit of the numerator and the denominator.
Step 3: For the numerator, apply the Sum/Difference Rule for limits: lim_{x→a} [u(x) ± v(x)] = lim_{x→a} u(x) ± lim_{x→a} v(x). This lets us separate the terms 2f(x) and -g(x) into individual limits.
Step 4: For the term 2f(x), apply the Constant Multiple Rule: lim_{x→a} [c * u(x)] = c * lim_{x→a} u(x). This allows us to take the constant 2 out of the limit.
Step 5: For the denominator, apply the Sum Rule and the Constant Rule: lim_{x→a} [u(x) + c] = lim_{x→a} u(x) + c, where c is a constant. This simplifies the denominator to (lim_{x→0} f(x) + 7)².

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