Limits (x increasing without bounds)

  • Thread starter chaosblack
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  • #1
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Before we start, can you just explain how to start me off? I understand the idea of all limits, just a few tricky ones that were assigned for homework. (note: ALL limits are x --> infinity)

Homework Statement


lim [tex]^{}\frac{10}{x}[/tex]


lim [tex]\frac{x^{2}+a^{2}}{x^{3}+a^{3}}[/tex]


lim (1 - r[tex]^{x}[/tex]) , |r| < 1

Homework Equations





The Attempt at a Solution



For 10, I'm guessing a = a constant number, so the answer would be zero either way?
 
Last edited:

Answers and Replies

  • #2
rock.freak667
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Before we start, can you just explain how to start me off? I understand the idea of all limits, just a few tricky ones that were assigned for homework. (note: ALL limits are x --> infinity)

Homework Statement


lim 10[tex]^{}\frac{10}{x}[/tex]


lim [tex]\frac{x^{2}+a^{2}}{x^{3}+a^{3}}[/tex]


lim (1 - r[tex]^{x}[/tex]) , |r| < 1
I am hoping these are the questions if I am reading them correctly

[tex]\lim_{x\rightarrow \infty} \frac{10}{x}[/tex]

[tex]\lim_{x\rightarrow \infty} \frac{x^{2}+a^{2}}{x^{3}+a^{3}}[/tex]

[tex]\lim_{x\rightarrow \infty} (1 - r^x) [/tex] for [tex] |r|<1[/tex]

For the first one..as x gets bigger and bigger, what happens to 1/x ? If you don't know, check it on a calculator.

For the second one, when you have it in a fraction like that. First you check if the degree of the denominator is less than or equal to the degree of the numerator. If so, you need to divide it out. If not, what is usually done is you divide the numerator and denominator by the highest power of x and then take the limit.

For the third one, if |r|<1..what does that mean for values of r? What kind of numbers would r have to be for the modulus to be less than one?
 
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  • #3
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Yeah sorry, I edited the original questions (and know I know how to show the limit's now haha).

So then my answers for 10/x would be 0. the one with "a" as a variable would be 0. and then the one with the absolute would be 1?
 
  • #4
rock.freak667
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Yep...that seems to be correct.
But here is another way to look at the 2nd one.
As [itex]x \rightarrow \infty[/itex], in the numerator,[itex]x^2+a^2[/itex], a^2 becomes negligeable so that the numerator is basically [itex]x^2[/itex]. Similarly,the denominator becomes [itex]x^3[/itex] as [itex]a^3[/itex] becomes negligeable as [itex]x \rightarrow \infty[/itex]

So your expression really reduces to this

[tex]\lim_{x \rightarrow \infty}\frac{x^2}{x^3} =\lim_{x \rightarrow \infty}\frac{1}{x}[/tex]
 

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