Lim h approaches 0 sin(pi/3+h)-sin(pi/3)
To find the limit as h approaches 0 of sin(pi/3+h) - sin(pi/3), we can use the trigonometric identity:
sin(A + B) - sin(A) = 2cos((A + B + A) / 2) * sin((A + B - A) / 2)
In this case, A = pi/3 and B = h. Substituting these values into the identity, we have:
sin(pi/3 + h) - sin(pi/3) = 2cos((pi/3 + h + pi/3) / 2) * sin((pi/3 + h - pi/3) / 2)
Simplifying further:
= 2cos((2pi/3 + h) / 2) * sin(h/2)
= 2cos(pi/3 + h/2) * sin(h/2)
Using the double angle formula for cosine:
= 2cos(pi/6)cos(h/2) * sin(h/2)
= 2(sqrt(3)/2)cos(h/2) * sin(h/2)
= sqrt(3)cos(h/2) * sin(h/2)
Taking the limit as h approaches 0:
lim(h->0) sqrt(3)cos(h/2) * sin(h/2) = sqrt(3) * cos(0/2) * sin(0/2)
= sqrt(3) * cos(0) * sin(0)
= sqrt(3)(1)(0)
= 0
Therefore, the limit as h approaches 0 of sin(pi/3+h) - sin(pi/3) is 0.
Does the following make sense? Make corrections:
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To evaluate the limit:
lim h approaches 0 [sin(pi/3+h) - sin(pi/3)]
We can use the trigonometric identity for the difference of two sines:
sin(A) - sin(B) = 2 * cos((A + B)/2) * sin((A - B)/2)
In this case, A = pi/3+h and B = pi/3. Substituting these values into the identity, we have:
sin(pi/3+h) - sin(pi/3) = 2 * cos((pi/3+h + pi/3)/2) * sin((pi/3+h - pi/3)/2)
Simplifying the expression:
= 2 * cos((2pi/3+h)/2) * sin(h/2)
= 2 * cos(pi/3 + h/2) * sin(h/2)
Now, as h approaches 0, the angle h/2 approaches 0 as well. Using the fact that cos(pi/3) = 1/2, we have:
lim h approaches 0 [2 * cos(pi/3 + h/2) * sin(h/2)]
= 2 * cos(pi/3) * sin(0)
= 2 * (1/2) * 0
= 0
Therefore, the limit is equal to 0.
To evaluate the limit of this expression, we can use the concept of the limit definition of the derivative. The limit definition of the derivative is given by:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
In our case, f(x) = sin(x). Therefore, to find the limit as h approaches 0 of sin(pi/3 + h) - sin(pi/3), we can rewrite it using the limit definition of the derivative:
lim(h->0) [sin(pi/3 + h) - sin(pi/3)] / h
Now, simplifying the expression:
= lim(h->0) [sin(pi/3 + h) - sin(pi/3)] / h
= lim(h->0) [sin(pi/3)cos(h) + cos(pi/3)sin(h) - sin(pi/3)] / h
Since sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2, we can substitute these values into the expression:
= lim(h->0) [ (sqrt(3)/2 * cos(h)) + (1/2 * sin(h)) - (sqrt(3)/2) ] / h
Now, we can evaluate the limit by substituting h = 0 into the expression:
= [ (sqrt(3)/2 * cos(0)) + (1/2 * sin(0)) - (sqrt(3)/2) ] / 0
= [ (sqrt(3)/2 * 1) + (1/2 * 0) - (sqrt(3)/2) ] / 0
= [ (sqrt(3)/2 - sqrt(3)/2) ] / 0
= [ 0 ] / 0
At this point, we have an indeterminate form of 0/0. This means we need to use a different method to evaluate the limit. One way to do this is by using L'Hôpital's rule, which states that if we have an indeterminate form of 0/0 or infinity/infinity, we can take the derivative of the numerator and denominator and then evaluate the limit again.
Taking the derivative of the numerator and denominator, we have:
= d/dh [ (sqrt(3)/2 * cos(h)) + (1/2 * sin(h)) - (sqrt(3)/2) ] / d/dh [h]
= [ (-sqrt(3)/2 * sin(h)) + (1/2 * cos(h)) ] / 1
= -sqrt(3)/2 * sin(h) + 1/2 * cos(h)
Now, we can evaluate the limit again by substituting h = 0 into the expression:
= -sqrt(3)/2 * sin(0) + 1/2 * cos(0)
= -sqrt(3)/2 * 0 + 1/2 * 1
= 0 + 1/2
= 1/2
Therefore, the limit of sin(pi/3 + h) - sin(pi/3) as h approaches 0 is equal to 1/2.