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 limits in the given expressions, we can make use of L'Hôpital's Rule, which states that if we have an indeterminate form of the type 0/0 or ∞/∞, we can differentiate the numerator and denominator repeatedly until we obtain a limit that is not of this form. Let's apply this rule to each limit.
a) Compute limx → ∞ (xF(x))/(e^(x^2)):
To evaluate this limit, we will differentiate both the numerator and denominator repeatedly until we obtain a non-indeterminate form.
First, let's differentiate the numerator:
F(x) = ∫[0, x] e^(t^2) dt
Using the Fundamental Theorem of Calculus, we can write:
F'(x) = e^(x^2)
Now, let's differentiate the denominator:
g(x) = e^(x^2)
g'(x) = 2x * e^(x^2)
And now we can compute the limit:
limx → ∞ (xF(x))/(e^(x^2)) = limx → ∞ (F'(x))/(g'(x))
Since both the numerator and denominator are now non-indeterminate forms, we can plug in ∞ for x:
limx → ∞ (xF(x))/(e^(x^2)) = limx → ∞ (F'(x))/(g'(x)) = limx → ∞ (e^(x^2))/(2x * e^(x^2))
Simplifying, we get:
limx → ∞ (xF(x))/(e^(x^2)) = 1/2
Therefore, the limit of (xF(x))/(e^(x^2)) as x approaches infinity is 1/2.
b) Compute limx → 0 F(x)/(xe^(x^2)):
To evaluate this limit, we will differentiate both the numerator and denominator repeatedly until we obtain a non-indeterminate form.
First, let's differentiate the numerator:
F(x) = ∫[0, x] e^(t^2) dt
Using the Fundamental Theorem of Calculus, we can write:
F'(x) = e^(x^2)
Now, let's differentiate the denominator:
g(x) = xe^(x^2)
g'(x) = e^(x^2) + 2x^2 * e^(x^2)
And now we can compute the limit:
limx → 0 F(x)/(xe^(x^2)) = limx → 0 (F'(x))/(g'(x))
Since both the numerator and denominator are now non-indeterminate forms, we can plug in 0 for x:
limx → 0 F(x)/(xe^(x^2)) = limx → 0 (F'(x))/(g'(x)) = limx → 0 (e^(x^2))/(e^(x^2) + 2x^2 * e^(x^2))
Simplifying, we get:
limx → 0 F(x)/(xe^(x^2)) = 1/1 = 1
Therefore, the limit of F(x)/(xe^(x^2)) as x approaches 0 is 1.