How to solve these logarithmic functions?2. What is the values (if any) of the following? e^x as x--> - infinity
You should evaluate the limit of the function `e^x` under the given condition, such that:
`lim_(x->-oo) e^x = e^(-oo)`
Using negative power property yields:
`a^(-b) = 1/a^b`
Reasoning by analogy yields:
`e^(-oo) = 1/e^oo = 1/oo = 0`
Hence, evaluating the limit of the given function `e^x` , under the given conditionsm, yields `lim_(x->-oo) e^x = 0.`
The number e is a mathematical constant with an approximate value 2.71828. This is positive and real.
As e is positive and greater than 1, the value of e^x increases as x increases.
Use the relation `x^(-a) = 1/(x^a)`
= `lim_(x->oo) 1/e^x`
As x tends to infinity, the value of e^x also tends to infinity. Consequently, the value of 1/x tends to 0.
The limit `lim_(x->-oo) e^x = 0`