Calculus homework help- lim as x --> 5pi/4 f(x)=(sin(x)-cos(x))/(cos2(x))?

Answer 1

Did you mean for the denominator to by cos²x? If so, the answer is simply:
(-1/√2 + 1/√2) / (-1/2)² = 0 / (1/4) = 0
__________

If you meant for the denominator to be cos 2x then:

Lim as x --> 5π/4 f(x) = [(sin x) - (cos x)]/(cos 2x)

Since this is of the indeterminant form 0/0 L'Hospital's Rule applies. It states that the limit of the quotient is equal to the quotient of the derivative of the numerator divided by the derivative of the denominator.

Lim x --> 5π/4 f(x) = [(sin x) - (cos x)]/(cos²x)

= Lim x --> 5π/4 [(cos x) + (sin x)] / [-2(sin x)(cos x)]

(-1/√2 - 1/√2) / [-2(1/√2)(1/√2)] = -√2 / (-1) = √2

Answer 2

you ought to draw a circle and then a triangle based on the beginning place O. enable A be a element on the fringe of the circle and make bigger it to a element B mendacity exterior of the circle. enable the different element opposite of A an mendacity on the fringe of the circle be reported as C and the the element perpendicular to A reported as T. Draw it out, be conscious that the element of triangle OAT is below the international and the international is below the element of the bigger triangle, so (a million/2)sinx < (a million/2)x < (a million/2)tanx yet this is purely ture for 0 < x < pi/2. Simpify: sinx < x < tanx a million < x/sinx < a million/cosx So a million > sinx/x > cosx additionally be conscious that cosx = a million - 2sin^2(x/2) > a million - (a million/2)x^2 So, a million - (a million/2)x^2 < cosx < sinx/x < a million by ability of the sandwhich theorem, sinx/x -> a million as x ->0.