# How to solve these logarithmic functions?2. What is the values (if any) of the following? e^x as x--> - infinity

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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) e^x`

= `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`