Chapter 3.5, Problem 51E

To determine

To show:That $\underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=0$

Answer to Problem 51E

  $\begin{array}{l}\underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle \lim }}\frac{(\cosh -1)(\cosh +1)}{h(\cosh +1)}\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle \lim }}\frac{{\cos }^{2}h-1}{h(\cosh +1)}\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle \lim }}\frac{-{\sin }^{2}h}{h(\cosh +1)}\\ \underset{h\to 0}{{\displaystyle \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\lim }}(\frac{\sinh }{h}.\frac{-\sinh }{\cos h+1})\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle (\lim }}\frac{\sinh }{h})(\underset{h\to 0}{{\displaystyle \lim }}\frac{-\sinh }{\cos h+1})\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=1(\frac{-\sin 0}{\cos 0+1})=\frac{0}{1+1}=0\end{array}$

Explanation of Solution

Given information:Use analytic method to show that $\underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=0$ .

Calculation:Multiply the numerator and denominator $\cosh +1$ we get:

  $\begin{array}{l}\underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle \lim }}\frac{(\cosh -1)(\cosh +1)}{h(\cosh +1)}\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle \lim }}\frac{{\cos }^{2}h-1}{h(\cosh +1)}\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle \lim }}\frac{-{\sin }^{2}h}{h(\cosh +1)}\\ \underset{h\to 0}{{\displaystyle \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\lim }}(\frac{\sinh }{h}.\frac{-\sinh }{\cos h+1})\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=\underset{h\to 0}{{\displaystyle (\lim }}\frac{\sinh }{h})(\underset{h\to 0}{{\displaystyle \lim }}\frac{-\sinh }{\cos h+1})\\ \underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}=1(\frac{-\sin 0}{\cos 0+1})=\frac{0}{1+1}=0\end{array}$