Then you need a new calculator! e^x does not go anywhere near 1 as x goes to infinity.laura_a said:Homework Statement
I have done an integration and ended up with the result
[-c/2 * [e^(-2x)]] |^infinity_0 = 1
The solution is that c=2 so that means to me that e^(2x) must turn into minus 1 for it to equal 1... but I'm not sure.. I've got graphcalc so I've been staring at the graph and I figure that as x goes to infinity that e^x goes to 1... but not sure what to say when x goes to minus infinity?
The limit of e^x as x approaches infinity is infinity. This means that the value of e^x will continue to increase without bound as x gets larger and larger.
Near infinity, e^x increases without bound, while near -infinity, e^x approaches 0. This is because as x approaches -infinity, the exponent becomes infinitely negative, causing the value of e^x to approach 0.
The natural logarithm function is the inverse of the exponential function, so as e^x approaches infinity, ln(x) also approaches infinity. This means that e^x and ln(x) are asymptotically related near infinity.
The Taylor series expansion of e^x is 1 + x + (x^2)/2! + (x^3)/3! + ... , which shows that as x approaches infinity, the value of e^x becomes increasingly dominated by the first term, 1. This aligns with the behaviour of e^x near infinity, where it approaches infinity.
No, e^x is always a positive value. As x approaches infinity, e^x may become a very large positive number, but it will never be negative.