find L- Hospital
lim x-> inf
f(x)=1/e^(-x)
To find the limit as x approaches infinity of the function f(x) = 1/e^(-x), we can apply L'Hospital's rule. L'Hospital's rule is a method used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞, by using derivatives.
Here's how we can apply L'Hospital's rule to this problem:
First, we need to rewrite the function to make it suitable for applying the rule. In this case, we can rewrite f(x) as:
f(x) = e^x
Now we have a function that is differentiable, and its derivative is equal to itself:
f'(x) = e^x
Next, we calculate the derivative of the function in the numerator and the denominator. In this case, both the numerator and denominator are the same, so the derivative of e^x is still e^x.
Now, we have:
f'(x) = e^x
We can apply L'Hospital's rule by taking the limit as x approaches infinity of the ratio of the derivatives:
lim(x->inf) [f'(x)] / [g'(x)] = lim(x->inf) [e^x] / [e^x]
Since the numerator and denominator are the same, we can cancel them out:
lim(x->inf) [e^x] / [e^x] = 1
Therefore, the limit of f(x) = 1/e^(-x) as x approaches infinity is 1.