Question

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?.

Answer 1

See the explanation.

Explanation:

As `x to -oo, cosh x to oo and sinh x to -oo`.

The arithmetic mean `(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 -oo, cosh x to oo, sinh x to -oo and e^x/2 to 0.`