HallsofIvy said:That looks an awful lot like an exponential function to me. Take the logarithm of [itex]1+(x/n))^n and you have a limit, as n goes to infinity, of the form [itex]0*\infty[/itex]. That can be rewritten as so that it is of the form "0/0" and, even though here n is an integer, you can use L'Hopital's rule.
The expression (1+(x/n))^n represents the limit as n approaches infinity of the function (1+(x/n))^n. This means that as n gets larger and larger, the value of the function will approach a certain number.
To calculate the limit of (1+(x/n))^n as n approaches infinity, you can use the formula lim n[tex]\rightarrow[/tex][tex]\infty[/tex] (1+(x/n))[tex]^{n}[/tex] = e^x. This means that the limit is equal to the value of the exponential function e^x, where x is the value inside the parentheses.
The limit (1+(x/n))^n as n approaches infinity is significant because it represents the behavior of the function as the input value (n) gets larger and larger. This is known as the limit at infinity and can help us understand the long-term behavior of a function.
Yes, the limit (1+(x/n))^n can be calculated for any value of x. This is because the limit is independent of the value of x and only depends on the behavior of the function as n approaches infinity.
The limit (1+(x/n))^n is closely related to the concept of exponential growth. As n approaches infinity, the value of the function (1+(x/n))^n will approach e^x, which is the value of exponential growth. This means that the function will grow exponentially as n gets larger, which is a common pattern in many real-world scenarios.