The graph of y=e^x/2 is the bisector-graph of the graphs for y = 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?.

Aug 21, 2016

See the explanation.

Explanation:

As $x \to - \infty , \cosh x \to \infty \mathmr{and} \sinh x \to - \infty$.

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

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