Another BC Calc limits question?

Question

What is the limit as h-->0 of [sin(pi/6 + h) - sin(pi/6)] / h ? It could help if you try writing it out in proper notation first.

Answer 1

limit as h-->0 of [sin(pi/6 + h) - sin(pi/6)] / h
=limit as h-->0 [sinpi/6cosh+cospi/6sinh -sinpi/6]/h
=limit as h-->0 [sinpi/6(cosh-1)+cospi/6sinh ]/h =0/0.
Since this is indeterminate, Use L'Hospital's Rule
=limit as h-->0 [0.5(cosh-1)+0.5sqrt(3)sinh ]/h
=limit as h-->0 [-0.5sinh+0.5sqrt(3)cosh ]/1 = 0.5sqrt(3)

Answer 2

Tailpipe had the correct idea but imperfectly drew the parallel. The definition of the derivative of sin(x) at the point (x,sin(x)) is limit h->0 of {sin(x+h)-sin(x)}/h. This derivative is cos(x). The problem then deals with the specific value x=pi/6. So the answer is cos(pi/6) which is the adjacent side to the small angle pi/6=30 degrees, so is equal to sqrt(3)/2. In other words Tailpipe got the words in the wrong order: it should've been the derivative of sin(x) then evaluated at x=pi/6 but he figured it was the derivative of the value sin(pi/6). If you are allowed a calculator on the test, you can check your answers during the test by using decimals. In other words, approximate h->0 by using h=0.00001 or something small like that. Get an approximate value for the limit and compare with your algebraic value from the theory.

Answer 3

Find the limit as h→0 of [sin(π/6 + h) - sin(π/6)] / h

Since both numerator and denominator are zero when h = 0, the quotient is indeterminant. Therefore use L'Hospital's rule. The limit of the derivative of the numerator and derivative of the denominator is the same as the limit of the original quotient.

limit as h→0 of [sin(π/6 + h) - sin(π/6)] / h
= limit as h→0 of [cos(π/6 + h) - 0] / 1
= cos(π/6) = (√3)/2

Answer 4

Its true that this is a constant, but 0? Wouldn't the limit jsut be sin (pi/6)?

Answer 5

This is the same as asking what is the derivative of sin(pi/6), which is 0 (it's a constant).

Answer 6

The best way is to use L'Hospital's rule:

lim f(x)/g(x) as x->0 = lim f'(x)/g'(x)

So here f(h) = sin(pi/6+h)-sin(pi/6)
and g(h) = h

So f'(h) = cos(pi/6+h)
and g'(h) = 1

So lim f(h)/g(h) = lim f'(h)/g'(h) = f(0)/1 = cos(pi/6)

So lim [sin(pi/6+h)+sin(pi/6)]/h = cos(pi/6) = sqrt(3)/2

Answer 7

First answer is right. 0

Answer 8

HUH?