Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Chapter Review - Review Exercises - Page 925: 11



Work Step by Step

Recall the limit rules: $ \ Rule -\ 1 \ : \lim\limits_{x \to a} [A(x)]^n =[\lim\limits_{x \to a} A(x)]^n $ $\ Rule \ - \ 2 : \lim\limits_{x \to a} A(x)=A(a)$ $\ Rule \ - \ 3: \lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$ $\lim\limits_{x\to 3}\dfrac{x^4-3x^3+x-3}{x^3-3x^2+2x-6}\\=\lim\limits_{x\to 3} \dfrac{x^3(x-3)+1(x-3)}{x^2(x-3)+2(x-3)}$ Apply $\lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$ $\lim\limits_{x\to 3}\dfrac{(x-3)(x^3+1)}{(x-3)(x^2+2)}=\lim\limits_{x\to 3}\dfrac{(x^3+1)}{(x^2+2)}\\=\dfrac{27+1}{9+2} \\=\dfrac{28}{11}$
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