Algebra.Com's Answer #609670 by jim_thompson5910(35256)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Let's focus on limit i. ![](\"/cgi-bin/rendertex.cgi?\LARGE) \r
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document.write( "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.\r
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document.write( "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).\r
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document.write( "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  \r
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document.write( "So the numerator of both limits is positive for t values near t = 4, but t is not actually equal to 4.\r
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document.write( "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)\r
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document.write( "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.\r
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document.write( "The one-sided limit will result in some negative number if  (we approach 4 from the left side)\r
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document.write( "Now onto limit ii. \r
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document.write( "The same idea applies from above. is never negative (see reasoning above).\r
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document.write( "The denominator is always positive if  (eg: if t = 4.5, then t-4 = 4.5-4 = +0.5)\r
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document.write( "So positive/positive = positive tells us that will result in some positive number if  (we approach 4 from the right side)\r
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document.write( "(b)\r
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document.write( "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.\r
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document.write( "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\r
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document.write( "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.\r
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document.write( "![](\"http://abcalculus.wikispaces.com/file/view/picture_of_a_cornerpoint.gif/52468863/279x174/picture_of_a_cornerpoint.gif\") \r
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document.write( "Note: the pieces of each half could be straight or curved.\r
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document.write( "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. \r
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document.write( "-------------------------------------------------------\r
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document.write( "If you need more help, or if you have any questions about the problem, feel free to email me at
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document.write( "jim_thompson5910@hotmail.com \n" );
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