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How would you find the limit of this?
Limit is as h approaches 0... the function is [cos(3pi/2 + h) - cos(3pi/2)] / h
I do notice that this does look like the definition of the derivative, but I do not know how to apply that knowledge to the problem.
2 Answers
- HemantLv 79 years agoFavorite Answer
This limit defines the Derivative of
cos x at x = 3π/2, i.e.
= [ ( -sin x ) at x = 3π/2 ]
= -sin (3π/2)
= -(-1)
= 1 ......................................... Ans.
______________________
- 9 years ago
Use the formula
cosA - cosB = 2[sin((B+A)/2) sin((B-A)/2)]
to write this as
2[sin(3pi/2 + h/2) sin(-h/2)]/h
= -2[sin(3pi/2 + h/2) sin(h/2)]/h
Since the numerator contains sin(h/2)
we want -h/2 in the denominator because of
lim[ x approaches 0] ((sin x)/x) = 1
so divide top and bottom by 2 to get
-1[sin(3pi/2 + h/2) sin(h/2)]/(h/2)
As h approaches 0, also h/2 approaches 0
therefore this expression approaches
-[sin(3pi/2 + 0)*1]
= - sin(3pi/2)
= 1.
You could also do this by recognising the limit of
[cos(x + h) - cos(x)] / h
as the derivative of cos x,
which is - sin x
then substitute x = 3pi/2
but I think it was not intended that you do it that way.
