The graph of $y = \frac{e^{x}}{2}$ $y=e^x/2$ is the bisector-graph of the graphs for $y = cosh x and y = sinh x$ $y = cosh x and y = sinh x$ . How do you use these graphs to show that the limit for the indeterminate form $\infty - \infty$ $oo-oo$ could be 0?.

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Answer 1 · 2016-08-21T17:13:43

See the explanation.

As $x \to - \infty, cosh x \to \infty and sinh x \to - \infty$ $x to -oo, cosh x to oo and sinh x to -oo$ .

The arithmetic mean $\frac{cosh x + sinh x}{2} = \frac{e^{x}}{2}$ $(cosh x + sinh x)/2 = e^x/2$ .

I have named the graph of #y =e^x/2 the bisector graph of the

graphs of y = cosh x and y = sinh x.

Use the separate limits:,

As $x \to - \infty, cosh x \to \infty, sinh x \to - \infty and \frac{e^{x}}{2} \to 0 .$ $x to -oo, cosh x to oo, sinh x to -oo and e^x/2 to 0.$

The graph of y=ex2y=e^x/2 is the bisector-graph of the graphs for y=coshxandy=sinhxy = cosh x and y = sinh x. How do you use these graphs to show that the limit for the indeterminate form ∞−∞oo-oo could be 0?.