Question 989301: The height of a ball above the ground at time t is h(t), and the ball bounces on a concrete floor at time t = 4 (seconds). Consider the following two one-sided limits:
i. lim t->4^- 
ii. lim t->4^+
(a). Which of the one-sided limits should be negative, and which positive? Explain your answer in a complete sentence.
(b). What would it mean physically for the absolute value of the two limits to be the same?
Please explain these in detail I am very confused.
Thank you
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! (a)
Let's focus on limit i. -h(4)}{t-4}\right))
It says "the ball bounces on a concrete floor at time t = 4 (seconds)". This means h(4) = 0. Basically, the height of the ball, h(t), is 0 when t = 4 seconds.
Anything near t = 4 is going to have h(t) be a positive value. It is impossible to have a negative height. The ball is in either in the air (height is positive) or it is on the ground (height is 0).
So again,  for any value of t. Since  for t values near t = 4, but t is not actually equal to 4, this makes  some positive number. Why? Because h(4) = 0, so 
So the numerator of both limits is positive for t values near t = 4, but t is not actually equal to 4.
The denominator  will be negative. If t was getting closer to 4 from the left side, then t would take on values such as t = 3, t = 3.5, t = 3.9, t = 3.99, t = 3.999, t = 3.9999, etc. We're getting closer to 4, but not actually getting there. We are approaching from the left side of 4. No matter which value of t you pick, the expression will be negative (eg: if t = 3.5, then t-4=3.5-4=-0.5)
Put this together and you'll find that  is overall a negative number if t was some value less than 4. We have positive/negative = negative in a sense.
The one-sided limit will result in some negative number if  (we approach 4 from the left side)
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Now onto limit ii. -h(4)}{t-4}\right))
The same idea applies from above. is never negative (see reasoning above).
The denominator is always positive if  (eg: if t = 4.5, then t-4 = 4.5-4 = +0.5)
So positive/positive = positive tells us that will result in some positive number if  (we approach 4 from the right side)
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(b)
The expression represents the average rate of change. Basically the average speed from (4,h(4)) to (t,h(t)). As t gets closer to 4, the average rate of change gets closer to the instantaneous rate of change. This is equal to the slope of the tangent line.
Let's say hypothetically that the left hand limit (LHL) was equal to -2. Also, let's say hypothetically that the right hand limit (RHL) was equal to +2
Since LHL does not equal RHL, the limit as t approaches 4 does not exist. However, notice how |-2| = |+2| = 2. This symmetry means we have a sharp point or cusp as you see below. This is one possible way the graph of h(t) could look like near t = 4.
Note: the pieces of each half could be straight or curved.
Another note: this is not the path the ball takes when it bounces. Keep in mind that the x axis is the time axis. The y axis is the height.
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If you need more help, or if you have any questions about the problem, feel free to email me at
jim_thompson5910@hotmail.com
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