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Christos Group Title

Limits, http://screencast.com/t/MeamUq9EgG8h I heard there is a rule about "powers" when it comes to limits like this in order to solve them? Can you tell me the rule please?

  • one year ago
  • one year ago

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  1. UnkleRhaukus Group Title
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    Is this your question ?\[\large\lim_{x\to+\infty}\frac{3x+1}{2x-5}\]

    • one year ago
  2. Christos Group Title
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    yes

    • one year ago
  3. zepdrix Group Title
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    Yes there is a nice little shortcut, which is what you must be referring to. I'm trying to think of how to explain it. Since it's an infinite limit, we only need to worry about the `leading` term (highest power on top and bottom). Compare them, if they're equal degree, the limit will approach the ratio of their coefficients. In this case, 3/2.

    • one year ago
  4. UnkleRhaukus Group Title
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    when \(x\to+\infty\), the ratio will ok like \(\frac{3x}{2x}\)

    • one year ago
  5. Christos Group Title
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    you mean like 3/2 ??

    • one year ago
  6. Christos Group Title
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    and when x->-inf -3/2 ?

    • one year ago
  7. zepdrix Group Title
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    Both x's are approaching negative infinity, so you can think of the negatives cancelling each other out, giving us the same 3/2 as with before :)

    • one year ago
  8. Christos Group Title
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    What about higher power at the denominator or the numerator separately

    • one year ago
  9. Christos Group Title
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    @zepdrix ?

    • one year ago
  10. zepdrix Group Title
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    It might help if you see an example and just think about it logically. \[\large \lim_{x \rightarrow \infty}\frac{\color{green}{x^2-2x+3}}{\color{royalblue}{x-7}}\] The green part represents a parabola, while the blue part a line. Which one will grow towards infinity faster? A straight line or a parabola?

    • one year ago
  11. Christos Group Title
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    1 ?

    • one year ago
  12. zepdrix Group Title
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    what? XD lol

    • one year ago
  13. Christos Group Title
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    uhm I think its the whole numerator? I am not sure :D

    • one year ago
  14. zepdrix Group Title
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    noooo it's not :O See how the `power` on x is larger in the top? We can't apply the same rule we used before, since the highest powers of x `do not match`.

    • one year ago
  15. zepdrix Group Title
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    |dw:1369665937413:dw|Which one of these functions is growing upwards faster? :O

    • one year ago
  16. zepdrix Group Title
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    The curve or the straight line? :D

    • one year ago
  17. Christos Group Title
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    numerator = denominator?

    • one year ago
  18. Christos Group Title
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    Eeeh i mean that high one not the straight one the other one grows faster

    • one year ago
  19. zepdrix Group Title
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    Yes good good, the curvy one grows faster.

    • one year ago
  20. zepdrix Group Title
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    \[\large \lim_{x \rightarrow \infty}\frac{\color{green}{x^2-2x+3}}{\color{royalblue}{x-7}}\]The `green` part is THAT type of function, the curvy one. So we can think of it this way, as the numerator and denominator approach infinity, the numerator is "winning", it's getting there faster. So the limit will approach infinity.

    • one year ago
  21. zepdrix Group Title
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    That too confusing? :3 stick with the math rules?

    • one year ago
  22. Christos Group Title
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    same applies for x->-inf ? and what about the denominator having the highest power?

    • one year ago
  23. zepdrix Group Title
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    Let's look at the reverse a moment. \[\large \lim_{x \rightarrow \infty}\frac{\color{royalblue}{x-7}}{\color{green}{x^2-2x+3}}\] Now the BOTTOM is a larger degree than the top. So the bottom is "winning", approaching infinity faster. So our limit will approach \(\large \dfrac{1}{\infty}\) which is the same as saying, it's approaching zero. The denominator is getting bigger and bigger faster than the top, so the fraction becomes smaller and smaller.

    • one year ago
  24. Christos Group Title
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    thanks

    • one year ago
  25. zepdrix Group Title
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    Negative infinity is a little trickier, since even powers will mess things up sometimes... I'm not sure how to explain that part properly :3

    • one year ago
  26. Christos Group Title
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    Just got back from my final exams on Calculus 1, 95% + :D .. I still can't believe that 3 months earlier I couldn't even do algebra. I got submitted to the University 3 weeks after the 1st semester was taking place.. They said no way a student with 10/20 overall score in high-school Algebra can take Calculus 1 from the very first semester, starting the course 3 WEEKS later.. I took this as a challenge, studied Algebra for additional 3 weeks straight. 6 weeks had past since the course of Calc had started at the uni. I went to the advisor's office, and said "I want to start Calculus 1. Give me the admission's test, if I pass, you let me in" . It was an algebra based test, scored 25/30 on it. Qualifying score to start studying Calculus 1 immediately. Teachers still doubting me, my score was pretty high on algebra they said, but starting Calculus 1 6 weeks after the lessons were taking place, I had 10% chance of even PASSING the course, they said. And here I am today, with OpenStudy as my friend and teacher, during this 3 months I studied algebra , qualified to take one of the most advanced forms of math Calculus 1 , scored 95%+ One of the highest if not the highest score in my class. Never in my life I thought that there would come a point where I would say "I love Math". A very big Thank You to the OpenStudy Community.

    • one year ago
  27. reemii Group Title
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    Congrats!

    • one year ago
  28. UnkleRhaukus Group Title
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    Well done @Christos ! \(\LARGE\mathcal{CONGRATULATIONS!!}\)

    • one year ago
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