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: 36

Answer

16

Work Step by Step

We want to find $\lim\limits_{x \to 4}\dfrac{4x-x^2}{2-\sqrt{x}}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 4 is zero, and if we try to substitute $x=4$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{4x-x^2}{2-\sqrt{x}}=\dfrac{x(4-x)}{2-\sqrt{x}}=\dfrac{x(2-\sqrt{x})(2+\sqrt{x})}{2-\sqrt{x}}=x(2+\sqrt{x})$ for all $x \neq 4.$ Thus $\lim\limits_{x \to 4}\dfrac{4x-x^2}{2-\sqrt{x}}=\lim\limits_{x \to 4}x(2+\sqrt{x})=4(2+\sqrt{4})=4(2+2)=4(4)=16.$
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