Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 52

Answer

(a) Quotient Rule. (b) Numerator: Root Rule - Denominator: Product Rule (c) Numerator: Constant Multiple Rule - Denominator: Difference Rule

Work Step by Step

(a) $\lim_{x\to1}\frac{\sqrt{5h(x)}}{p(x)(4-r(x))}=\dfrac{\lim_{x\to1}\sqrt{5h(x)}}{\lim_{x\to1}(p(x)(4-r(x)))}$ Here, $\lim_{x\to c}\dfrac{p}{q}=\dfrac{\lim_{x\to c}p}{\lim_{x\to c}q}$. The Quotient Rule is applied. (b) In the numerator we have $\lim_{x\to c}\sqrt[n]p=\sqrt[n]{\lim_{x\to c}p}$. The Root Rule is applied. and in the denominator: $\lim_{x\to c}(pq)=\lim\limits_{x\to c}p\times\lim\limits_{x\to c}q$. The Product Rule is applied. (c) Here, in the numerator: $\lim_{x\to c}(np)=n\lim_{x\to c}p$. The Constant Multiple Rule is applied and in the denominator we have $\lim_{x\to1}(4-r(x))=\lim_{x\to1}4-\lim_{x\to1}r(x)$. The Difference Rule is applied.
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