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course Mth 163
3-15-13 1:02
Question: Explain why the negative y axis is an asymptote for the function y = log{base 2}(x). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Your solution:
The range of an exponential growth problem is zero to infinity, so the domain of the inverse logarithmic problem is zero to infinity.
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This is a good answer, assuming we know why the range of the exponential function is zero to infinity, and assuming we know the limit zero to occur as x approaches -infinity.
To establish this you have to show that, as x approaches -infinity, 2^x approaches zero
I do want one revision. I want you to give me your best argument supporting the assertion that as x approaches -infinity, 2^x approaches zero. No calculator result can be used to support a valid argument. Don't spend an inordinate amount of time on this question. Limit yourself to at most 10 or 15 minutes then tell me what you're thinking.
You can just send me a copy of this note and your argument
As x approaches negative infinity, 2^x approaches zero because decimal gets smaller and smaller. The limit as x approaches infinity is 1/ 2^x. The chart below shows this:
X y=2^x
1 ½
3 .125
5 .03125
-10 1/x^10 1 / 2^10, but I know that's what you meant
-100 1/ 2^100 This is such a large decimal number that we say it will never reach zero
-1000 1/ 2^1000
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Good.
Another way to argue:
No matter how small you want 1/ 2^x to be, you can make it that small.
Let epsilon stand for a small positive number, as small as you might wish.
Then 1 / epsilon is a large positive number.
If 2^x is larger than 1 / epsilon, then 1 / 2^x is smaller than epsilon.
So all we need to do is show that there is some number x such that 2^x is bigger than 1 / epsilon.
Since 2^x doubles everytime x increases by 1, there is no limit to how big 2^x can get. So no matter how small the number epsilon and how large 1 / epsilon, 2^x will eventually exceed it.
Furthermore this will be so for all larger values of x.
It follows that from this value of x on, 1 / 2^x will be smaller than epsilon.
This is true for any epsilon, so there is no limit to how small 1 / 2^x can get.
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