Question
Lim h approaches 0 sin(pi/3+h)-sin(pi/3)
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Hello! My name is Pedro Torres. I am reaching out today because I have a question. I am currently advertising multiple amazon kdp books on amazon. c o m However, I setted daily budget for each one of my books that are beind advertise. Some are being setted to $5 dollars and some to $6 dollars. However, I been noticing that 4 of them are surpassing the amount. For example, one of them was suppose to stop at $5 dollars but it has reach to $10 dollars in one day. I would kindly like to know why is this occurring?
Hello! My name is Pedro Torres. I am reaching out today because I have a question. I am currently advertising multiple amazon kdp books on amazon. c o m However, I setted daily budget for each one of my books that are beind advertise. Some are being setted to $5 dollars and some to $6 dollars. However, I been noticing that 4 of them are surpassing the amount. For example, one of them was suppose to stop at $5 dollars but it has reach to $10 dollars in one day. I would kindly like to know why is this occurring?
Answered by
GPT 3.5
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.
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.
Answered by
GPT 3.5
Hello! My name is Pedro Torres. I am reaching out today because I have a question. I am currently advertising multiple Amazon KDP books on Amazon.com. However, I have set a daily budget for each of my books that are being advertised. Some are set at $5 and others at $6. However, I have noticed that 4 of them have exceeded the allocated budget. For example, one of them was supposed to stop at $5, but it has reached $10 in just one day. I would like to kindly know why this is happening.
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