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anonymous

  • one year ago

Simple question; Asymptotes with two answers - need explanation (look below for image and more details)

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  1. anonymous
    • one year ago
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    I know the solution is 6/2 and 3 , however I dont understand why -3 turns into 3. Would appreciate explaination :)

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  2. anonymous
    • one year ago
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    hey are you still here, I can help but I would need you to participate

  3. Jhannybean
    • one year ago
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    for \(\lim_{x\rightarrow \infty} \dfrac{6x^2-3x+4}{2x^2-8}\) just compare the coefficients of the highest power, in this case, \(\color{red}{x^2}\)

  4. Jhannybean
    • one year ago
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    That will give you your reason for the limit being 3.

  5. anonymous
    • one year ago
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    \[\frac{ 6x^{2}-3x + 4 }{ 2x^{2}-8 } = \frac{ x^{2}(6 - \frac{ 3 }{ x }+ \frac{ 4 }{ x^{2} }) }{ x^{2}(2-\frac{ 8 }{ x^{2} } )} = \frac{ 6 - \frac{ 3 }{ x } +\frac{ 4 }{ x^{2} }}{ 2-\frac{ 8 }{ x^{2} } }\] From here if you take the limit it should make sense how you get a limit of 3. \[\lim_{x \rightarrow \infty}\frac{ 6 - \frac{3}{x} + \frac{4}{x^{2}} }{ 2-\frac{8}{x^{2}} } = \frac{ 6 - 0 + 0 }{ 2- 0 } = 3\] Now on the 2nd line where you started doing the limit a different way, it looked like you divided every term by x instead of \(x^{2}\). Of course dividing by (or factoring out) \(x^{2}\) is what should be done, but let me take it from where you had it with only dividing by x: (after dividing by x) \[\lim_{x \rightarrow \infty}\frac{ 6x -3 + \frac{4}{x} }{ 2x - \frac{8}{x} } = \frac{ \infty - 3 + 0 }{ \infty - 0 }\] When you got to here, it looked like you tried to cancel out infinity or something. First of all, you cannot cancel out infinity. Different parts of a function can go to plus or minus infinity at different speeds. There is no equality from one infinity to another, they cannot cancel out. \[\frac{ \infty - 3 + 0 }{ \infty - 0 } = \frac{ \infty }{ \infty }\] This is an indeterminant form. In which case, if you wanted although its unncessary, you could use L' hopital's rule but the above shows why that isnt needed. So that was your error, cannot cross out infinity and the final result should be the indeterminant form \(\frac{\infty}{\infty}\). Long explanation, but hopefully that makes sense.

  6. anonymous
    • one year ago
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    wow @Concentrationalizing that was a pretty dang good response man...

  7. anonymous
    • one year ago
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    Thanks :)

  8. anonymous
    • one year ago
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    It's only good if it makes sense though, lol.

  9. sepeario
    • one year ago
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    I think you should make sure you consider infinity as a concept, not a number.

  10. anonymous
    • one year ago
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    Oh my god, I am an idiot! Of COURSE 6/2 is 3!! I was reading some examples and got the impression there would be two separate answers! But it was one all along.. Thanks a lot! *awards 10 dumb points to self*

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