Let F(x) = 0∫x e^(t^2)dt
a) Compute limx → ∞ (xF(x))/ (e^(x^2))
b) Compute limx → 0 F(x)/(xe^(x^2))
To compute the given limits, we can use L'Hopital's Rule, which states that for functions f(x) and g(x) whose derivatives exist on an interval containing a point c (except possibly at c), if the limit of f(x)/g(x) as x approaches c is of an indeterminate form, such as 0/0 or ∞/∞, then the limit of their derivatives f'(x)/g'(x) as x approaches c is equal to the original limit.
a) Compute limx → ∞ (xF(x))/ (e^(x^2))
To compute this limit, we need to find the derivatives of both the numerator and denominator and evaluate the limit of their ratio. Let's start by finding the derivative of F(x).
F(x) = 0∫x e^(t^2) dt
Using the Fundamental Theorem of Calculus, we can find the derivative of F(x) by applying the chain rule:
F'(x) = e^(x^2)
Next, let's find the derivative of e^(x^2):
(e^(x^2))' = 2x * e^(x^2)
Now we can evaluate the limit by applying L'Hopital's Rule:
limx → ∞ (xF(x))/ (e^(x^2)) = limx → ∞ (x * e^(x^2))/(2x * e^(x^2))
Canceling out the common factor of x and e^(x^2):
limx → ∞ 1/2 = 1/2
Therefore, the limit is 1/2.
b) Compute limx → 0 F(x)/(xe^(x^2))
To compute this limit, we'll need to find the derivative of the numerator and denominator and evaluate the limit of their ratio. Let's start by finding the derivative of F(x) again:
F'(x) = e^(x^2)
Next, let's find the derivative of xe^(x^2):
(xe^(x^2))' = e^(x^2) + 2x^2 * e^(x^2)
Now we can evaluate the limit by applying L'Hopital's Rule:
limx → 0 F(x)/(xe^(x^2)) = limx → 0 (e^(x^2))/(e^(x^2) + 2x^2 * e^(x^2))
Canceling out the common factor of e^(x^2):
limx → 0 1/(1 + 2x^2) = 1/1 = 1
Therefore, the limit is 1.