Q. 15 The figure that follows at the l... [FREE SOLUTION]

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Chapter 2: Q. 15 (page 233)

The figure that follows at the left shows the graphs of $y = \sinh x, y = \frac{1}{2} e^{x}, and y = - \frac{1}{2} e^{- x}$ .For each of the statements that follows, explain graphically why the statement is true. Then justify the statement algebraically, using the definitions of the hyperbolic functions.
$(a) \sinh x = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} (b) \lim_{x 鈫� - 鈭�} \frac{\sinh x}{- (1 / 2) e^{- x}} = 1$

Short Answer

Expert verified

Hence proved.

Step by step solution

Part (a) Step 1. Given information.

The given equation is,

$\sinh x = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x}$

Part (a) Step 2. Explanation.

$\sinh x = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} \frac{e^{x} - e^{- x}}{2} = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x}$

Hence Proved.

Part (b) Step 1. Explanation.

Taking the limits,

$\lim_{x 鈫� - 鈭�} \frac{\sinh x}{- \frac{1}{2} e^{- x}} = - \lim_{x 鈫� - 鈭�} \frac{\frac{\frac{1}{2} e^{x} - \frac{1}{2} e^{- x}}{e^{- x}}}{\frac{\frac{1}{2} e^{- x}}{e^{- x}}} = - \lim_{x 鈫� - 鈭�} \frac{\frac{e^{x} - e^{- x}}{e^{- x}}}{\frac{e^{- x}}{e^{- x}}} = - \lim_{x 鈫� - 鈭�} \frac{\frac{e^{x} - e^{- x}}{e^{- x}}}{1} = - \lim_{x 鈫� - 鈭�} (e^{2 x} - 1) \lim_{x 鈫� - 鈭�} \frac{\sinh x}{- \frac{1}{2} e^{- x}} = 1$