Explain the following statements:
a. A jump NEVER has a limit
b. An asymptote (infinite) might have a limit
c. A removable ALWAYS has a limit
Monday, August 30, 2010
Sunday, August 29, 2010
Lindsey's Reflection
This is our first blog of the year..
Senior year at that.
it is crazy.
the first week of school brob was absent a couple of days and during that time we did problems
dealing with the quadratic formula.
that was mostly a review and it was not hard.
When she got back we went over the summer packet except for limits.
We spent the next week and a half going over limits.
we learned that for most of the time you can plug in for whatever problem you had.
an example would be
lim 2x+4
x-->2
for this problem it is not division or anything to where you would get in the bottom of the problem.
there fore you can easily plug in for this equation and get an acceptable answer.
lim 2x+4
x-->2
you would plug the 2 in where the x is to get the answer.
2(2)+4
4+4=8
the limit for this problem would be 8..
this for me is the easiest type of limit problem to solve but
unfortunately each time will not be as easy as the next.
in some problem you have to plug the problem into a table and find your answer that way.
this is the graph.
x -.1 -.01 -.001 .001 .01 .1
this is what you plug the problem into.
you have to add the number given to all of the numbers.
after doing so you plug the whole equation in to your graphing calculator and the table that you have set up will give you the answers for each different number that you have figured in the table.
most of the time both sides will be approaching a specific number and that will end up being the limit of the problem. you have to compare both sides of the problem and this will only work if both sides that you have compared are the same or they match each other.
in the case that the right and the left side are not the same or do not match..
then the limit does not exist and this will be the answer of the problem!
see everyone tommorrow!
Senior year at that.
it is crazy.
the first week of school brob was absent a couple of days and during that time we did problems
dealing with the quadratic formula.
that was mostly a review and it was not hard.
When she got back we went over the summer packet except for limits.
We spent the next week and a half going over limits.
we learned that for most of the time you can plug in for whatever problem you had.
an example would be
lim 2x+4
x-->2
for this problem it is not division or anything to where you would get in the bottom of the problem.
there fore you can easily plug in for this equation and get an acceptable answer.
lim 2x+4
x-->2
you would plug the 2 in where the x is to get the answer.
2(2)+4
4+4=8
the limit for this problem would be 8..
this for me is the easiest type of limit problem to solve but
unfortunately each time will not be as easy as the next.
in some problem you have to plug the problem into a table and find your answer that way.
this is the graph.
x -.1 -.01 -.001 .001 .01 .1
this is what you plug the problem into.
you have to add the number given to all of the numbers.
after doing so you plug the whole equation in to your graphing calculator and the table that you have set up will give you the answers for each different number that you have figured in the table.
most of the time both sides will be approaching a specific number and that will end up being the limit of the problem. you have to compare both sides of the problem and this will only work if both sides that you have compared are the same or they match each other.
in the case that the right and the left side are not the same or do not match..
then the limit does not exist and this will be the answer of the problem!
see everyone tommorrow!
Reflection 1
One thing that I grasped from this week was the “Strategies” that were given to us.
They are:
1) Factor and cancel
2) Rationalize (for square roots)
3) Graph or chart
Example: Factor and cancel
Lim as x approaches 2 X^2 – 7x + 10/x^2 – 4
-First you must factor, and you get: (x – 5) (x – 2)/(x – 2) (x + 2)
-Now you cancel the values from the numerator and denominator that are the same:
Leaving you with (x – 5)/(x + 2)
-Now plug in your x value (2 because it is the limit as x approaches 2) and you get -3/4.
Example: Rationalizing
Lim as x approaches 25 5 – sq. rt. (x)/ 25 – x
-When rationalizing, you must multiply the top and bottom by the conjugate, which in this problem is 5 + sq.rt. (x).
-After doing this, you get: 25 – x/(25 – x) (5 + sq.rt. (x))
-Now cancel the 25 – x’s and you are left with 1/5 + sq.rt. (x)
-Now plug in x, and you get 1/10.
Graph or chart
When possible, you can use your calculator to either graph out the equation and find the limit visually or make the chart we learned at the beginning of the year, and you can find the limit that way.
THE KEY TO THIS is just practicing. All of the problems we will have will be similar, you just have to practice them until you get a complete understanding of how to work them out, not to mention practice will help you to do the process much faster.
They are:
1) Factor and cancel
2) Rationalize (for square roots)
3) Graph or chart
Example: Factor and cancel
Lim as x approaches 2 X^2 – 7x + 10/x^2 – 4
-First you must factor, and you get: (x – 5) (x – 2)/(x – 2) (x + 2)
-Now you cancel the values from the numerator and denominator that are the same:
Leaving you with (x – 5)/(x + 2)
-Now plug in your x value (2 because it is the limit as x approaches 2) and you get -3/4.
Example: Rationalizing
Lim as x approaches 25 5 – sq. rt. (x)/ 25 – x
-When rationalizing, you must multiply the top and bottom by the conjugate, which in this problem is 5 + sq.rt. (x).
-After doing this, you get: 25 – x/(25 – x) (5 + sq.rt. (x))
-Now cancel the 25 – x’s and you are left with 1/5 + sq.rt. (x)
-Now plug in x, and you get 1/10.
Graph or chart
When possible, you can use your calculator to either graph out the equation and find the limit visually or make the chart we learned at the beginning of the year, and you can find the limit that way.
THE KEY TO THIS is just practicing. All of the problems we will have will be similar, you just have to practice them until you get a complete understanding of how to work them out, not to mention practice will help you to do the process much faster.
Friday, August 27, 2010
Reflection
This week in calculus, we started reviewing and learning some new things about limits. A limit is a y-value. Whenever you are supposed to find a limit numerically, you use the table and if the numbers don’t match then the answer does not exist (DNE).
For Example:
Lim sin 2x/x
x>0
First you subtract or add 0 from -.1, -.01, -.001, .001, .01, and .1 to find the x-value.
Then you plug into your calculator for the y-value and should get 1.9867, 1.9999, 2, 2, 1.9999, and 1.986.
The means the limit is 2 because each side is approaching the same number.
Also, we learned how to plug in the number to find a limit.
For Example:
Lim x^4
x>-2
For this one you would plug in -2 for x, which would give you -2^4, so the answer would be 16.
Another Example:
Lim 2x-3/x+5
x>1
In this problem you plug in 1 for x, which would give you -1/6.
For Example:
Lim sin 2x/x
x>0
First you subtract or add 0 from -.1, -.01, -.001, .001, .01, and .1 to find the x-value.
Then you plug into your calculator for the y-value and should get 1.9867, 1.9999, 2, 2, 1.9999, and 1.986.
The means the limit is 2 because each side is approaching the same number.
Also, we learned how to plug in the number to find a limit.
For Example:
Lim x^4
x>-2
For this one you would plug in -2 for x, which would give you -2^4, so the answer would be 16.
Another Example:
Lim 2x-3/x+5
x>1
In this problem you plug in 1 for x, which would give you -1/6.
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