Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 34

Answer

\[\lim_{x\rightarrow 0}\frac{\cos mx-\cos nx}{x^2}=\frac{n^2-m^2}{2}\]

Work Step by Step

Let \[l=\lim_{x\rightarrow 0}\frac{\cos mx-\cos nx}{x^2}\] Which is $\frac{0}{0}$ form Using L'Hopital's rule \[l=\lim_{x\rightarrow 0}\frac{(\cos mx-\cos nx)'}{(x^2)'}\] \[\Rightarrow l=\lim_{x\rightarrow 0}\frac{-m\sin mx+n\sin nx}{2x}\] Which is again a $\frac{0}{0}$ form Using L'Hopital's rule \[\Rightarrow l=\lim_{x\rightarrow 0}\frac{(-m\sin mx+n\sin nx)'}{(2x)'}\] \[\Rightarrow l=\lim_{x\rightarrow 0}\frac{-m^2\cos mx+n^2\cos nx}{2}\] \[\Rightarrow l=\frac{n^2-m^2}{2}\] Hence , \[\lim_{x\rightarrow 0}\frac{\cos mx-\cos nx}{x^2}=\frac{n^2-m^2}{2}\]
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