Limits As X Approaches Infinity and Negative Infinity


by dylmans
Tags: approaches, infinity, limits, negative
dylmans
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#1
Sep8-08, 10:59 PM
P: 13
1. The problem statement, all variables and given/known data
Find the limit of each function
(a) as x approaches infinity and
(b) as x approaches negative infinity


2. Relevant equations
1. g(x)=1/(2+(1/x))

2. f(x)=(2x+3)/(5x+7)

3. h(x)=(9x^4+x)/(2x^4+5x^2-x+6)


3. The attempt at a solution
I don't know where to start.
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Dick
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#2
Sep8-08, 11:10 PM
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Put a large positive and a large negative number into each. Does that suggest what the answer might be? Now can you figure out how to prove it? This typically means dividing the numerator and denominator by the dominant power.
dylmans
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#3
Sep8-08, 11:22 PM
P: 13
i dont exactly get what you mean. what number would i need to put in? answers for number 1 are both 1/2 and the answers for number 2 are both 2/5. the answer for 3 isn't in the book

Dick
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#4
Sep8-08, 11:29 PM
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Limits As X Approaches Infinity and Negative Infinity


I was just suggesting you experiment numerically to get a feel for what a limit means. Try x=-100000 and x=100000. Are the answers close to what the book said? Now just start with the first one. What the limit of 1/x as x goes to either infinity?
dylmans
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#5
Sep8-08, 11:31 PM
P: 13
1 over infinity?
Dick
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#6
Sep8-08, 11:38 PM
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Quote Quote by dylmans View Post
1 over infinity?
No, no. LIMIT 1/x as x goes to infinity. It APPROACHES an honest to God REAL number. What is that actual number? Infinity is not a number. 1/10, 1/100, 1/1000, 1/10000. What are they getting closer and closer to?
dylmans
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#7
Sep8-08, 11:44 PM
P: 13
they're getting closer to infinity...
Dick
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#8
Sep8-08, 11:47 PM
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No, son, they are getting closer to zero. Let's rewrite them in decimal 0.1, 0.01, 0.001, 0.0001, etc.
dylmans
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#9
Sep8-08, 11:55 PM
P: 13
oh duh, ok, so what do i need to do to figure out the answer, plug in zero?
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#10
Sep8-08, 11:58 PM
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Well, what's 1/(2+almost zero)? For the second one divide numerator and denominator by x. Now you have (2+almost zero)/(5+almost zero).
dylmans
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#11
Sep9-08, 12:06 AM
P: 13
ok, i think i get those two better. so for 3, id start by trying to factor, which i don't see anything that factors off the top of my head...
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#12
Sep9-08, 12:10 AM
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Divide numerator and denominator by x^4. As x->infinity, x^4 is the large term. The rest go to zero for the same reason 1/x did.
dylmans
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#13
Sep9-08, 12:24 AM
P: 13
ok, i think i get it now. so the answer for number 3 would be 9/2 as it approaches infinity and the same as it approaches negative infinity?
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#14
Sep9-08, 12:32 AM
P: 13
ok so the next problem is the limit as x approaches infinity for (2+x^1/2)/(2-x^1/2). the answer i got was -1.
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#15
Sep9-08, 12:36 AM
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I believe that.
dylmans
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#16
Sep9-08, 12:39 AM
P: 13
ok, that works, thanks for the help. i'll post if i run into anymore problems


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