91Ó°ÊÓ

We are given a graph of 3 functions which are$y=\tan h\left(x\right),y=1,andy=-1.$

(a) As we can see from the graph that the complete graph of y =tanh x lies between the graphs of $y=1andy=-1$. Therefore, $-1â\copyright ½\tan hxâ\copyright ½1.$

(b) Similarly from the graph as x increases and tends towards infinity the graph of tanh x moves towards 1 therefore, $\underset{x→∞}{\lim }\tan hx=1.$

Also, from the graph as x decreases and tends towards infinity the graph of tanh x moves towards -1 therefore,$\underset{x→-∞}{\lim }\tan hx=-1.$

$$\begin{gathered} Asweallknow, \\ \tan h\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}, \\ Solvingthisbydividingnumeratoranddenominatorby{e}^{x}weget, \\ \tan h\left(x\right)=\frac{1-e^{-2x}}{1+e^{-2x}}. \\ Nowwhenxtendstoinfinitythen{e}^{-2x}tendsto0and\tan h\left(x\right)tendsto1. \\ Similarlywhenxtendsto-infinitythen{e}^{-2x}tendsto∞and\tan h\left(x\right)tendsto-1. \\ Therefore,ontheentirenumberline\tan h\left(x\right)liesbetween1and-1. \end{gathered}$$

$$\begin{gathered} \underset{x→∞}{\lim }\tan h\left(x\right)=\underset{x→∞}{\lim }\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) \\ =\underset{x→∞}{\lim }\left(\frac{1-e^{-2x}}{1+e^{-2x}}\right) \\ =\underset{x→∞}{\lim }\left(\frac{1-0}{1-0}\right)=1. \\ \underset{x→-∞}{\lim }\tan h\left(x\right)=\underset{x→-∞}{\lim }\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) \\ =\underset{x→-∞}{\lim }\left(\frac{e^{2x}-1}{e^{2x}+1}\right) \\ =\underset{x→-∞}{\lim }\left(\frac{0-1}{0+1}\right)=-1. \\ Hence,Proved. \end{gathered}$$