is that because when x aproaches infinity cosh(x) = infinity and as e^x also approaches infinity when x approaches infinity it can be said that lim(x-> infinity) cosh(x) = (lim x-> infinity)e^x? Also, thanks for replying!phyzguy said:
thegirl said:
The limit of cosh and sinh is infinity. This means that as the input approaches infinity, the output of both functions will also approach infinity.
Cosh and sinh are two related hyperbolic functions. Cosh is the hyperbolic cosine function, while sinh is the hyperbolic sine function. The main difference between the two is that cosh is an even function, meaning it is symmetric about the y-axis, while sinh is an odd function, meaning it is symmetric about the origin.
To calculate the limit of cosh and sinh algebraically, you can use the fact that cosh(x) = (e^x + e^-x)/2 and sinh(x) = (e^x - e^-x)/2. As x approaches infinity, e^x will become much larger than e^-x, so you can simplify the expressions to e^x/2 and e^x/2, respectively. Since e^x approaches infinity as x approaches infinity, the limit of both cosh and sinh will also be infinity.
The graph of cosh and sinh are both hyperbolic functions, which means they have a similar shape to a parabola. However, while a parabola opens either upwards or downwards, the graph of cosh opens upwards and the graph of sinh opens downwards. Both graphs also have an asymptote at y=1, and approach infinity as x approaches infinity.
Cosh and sinh have various real-world applications, including in physics, engineering, and finance. They are often used to model the growth or decay of physical quantities, such as temperature or population. In finance, they are used to model the growth of investments or to calculate compound interest. They are also used in electrical engineering to model alternating current circuits.