Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 34

Answer

$$\lim_{x\to0}\frac{\cos mx-\cos nx}{x^2}=\frac{n^2-m^2}{2}$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\cos mx-\cos nx}{x^2}$$ For $\lim_{x\to0}(\cos mx-\cos nx)=\cos(0m)-\cos(0n)=\cos0-\cos0=0$ and $\lim_{x\to0}(x^2)=0^2=0,$ we have an indeterminate form of $\frac{0}{0}$, and we can apply L'Hospital's Rule here: $$A=\lim_{x\to0}\frac{(\cos mx-\cos nx)'}{(x^2)'}$$ $$A=\lim_{x\to0}\frac{-m\sin mx-(-n\sin nx)}{2x}$$ $$A=\lim_{x\to0}\frac{-m\sin mx+n\sin nx}{2x}.$$ $\lim_{x\to0}(-m\sin mx+n\sin nx)=-m\sin(0m)+n\sin(0n)=-m\sin0+n\sin0=-0m+0n=0$ and $\lim_{x\to0}(2x)=2\times0=0,$ which is another indeterminate form of $\frac{0}{0}$. Applying L'Hospital's Rule, we have $$A=\lim_{x\to0}\frac{(-m\sin mx)'+(n\sin nx)'}{(2x)'}$$ $$A=\lim_{x\to0}\frac{-m^2\cos mx+n^2\cos nx}{2}$$ $$A=\frac{-m^2\cos(0m)+n^2\cos(0n)}{2}$$ $$A=\frac{-m^2\cos0+n^2\cos0}{2}$$ $$A=\frac{-m^2\times1+n^2\times1}{2}$$ $$A=\frac{n^2-m^2}{2}$$
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