1) lim 1-cos x / x^2
x->0
2) use the mean value theorem to
determine the value of c, given that
f(x) = cos5x
1) To find the limit of the function (1 - cos x) / x^2 as x approaches 0, we can use some algebraic manipulations and L'Hôpital's Rule.
First, let's simplify the expression by using the formula cos(2θ) = 1 - 2sin^2(θ):
1 - cos x = 1 - (1 - 2sin^2(x/2)) = 2sin^2(x/2)
Now, let's rewrite the expression:
lim (1 - cos x) / x^2 = lim (2sin^2(x/2)) / x^2
Since this limit is in an indeterminate form of 0/0, we can apply L'Hôpital's Rule. By taking the derivatives of the numerator and denominator with respect to x, we have:
lim (2sin^2(x/2))' / (x^2)' = lim (4sin(x/2)cos(x/2)) / 2x
Next, substitute x = 0 into the expression:
lim (4sin(0/2)cos(0/2)) / (2*0) = lim (4sin(0)*cos(0)) / 0 = 0 / 0
This is still in an indeterminate form, so we can apply L'Hôpital's Rule again:
lim (4sin(x/2)cos(x/2))' / (2x)' = lim (4(0.5)cos(x/2)cos(x/2) - 4sin(x/2)sin(x/2)) / 2 = lim 2cos^2(x/2) - 2sin^2(x/2) / 2 = lim 2(cos^2(x/2) - sin^2(x/2)) / 2
Now, substitute x = 0 into the expression:
lim 2(cos^2(0/2) - sin^2(0/2)) / 2 = lim 2(cos^2(0) - sin^2(0)) / 2 = 2(1 - 0) / 2 = 2 / 2 = 1
Therefore, the limit of (1 - cos x) / x^2 as x approaches 0 is 1.
2) To apply the Mean Value Theorem to find the value of c, we need to satisfy the conditions of the theorem:
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the open interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
In this case, f(x) = cos(5x) and we need to find the value of c.
The first step is to find the values of f(a) and f(b). Since the interval is not specified, we can choose any values for a and b. Let's choose a = 0 and b = π/4:
f(a) = f(0) = cos(5 * 0) = cos(0) = 1
f(b) = f(π/4) = cos(5 * π/4) = cos(5π/4) = -1/sqrt(2)
Now we can find the derivative of f(x) using the chain rule:
f'(x) = d/dx [cos(5x)] = -5sin(5x)
Next, we calculate the difference quotient:
[f(b) - f(a)] / (b - a) = [-1/sqrt(2) - 1] / (π/4 - 0) = [-1/sqrt(2) - 1] / (π/4)
Now, set this equal to f'(c) and solve for c:
-5sin(5c) = [-1/sqrt(2) - 1] / (π/4)
By solving this equation numerically, we can find the value of c.