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Christos

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?

  • 11 months ago
  • 11 months ago

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

    • 11 months ago
  2. Christos
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    yes

    • 11 months ago
  3. zepdrix
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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.

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

    • 11 months ago
  5. Christos
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    you mean like 3/2 ??

    • 11 months ago
  6. Christos
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    and when x->-inf -3/2 ?

    • 11 months ago
  7. zepdrix
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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 :)

    • 11 months ago
  8. Christos
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    What about higher power at the denominator or the numerator separately

    • 11 months ago
  9. Christos
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    @zepdrix ?

    • 11 months ago
  10. zepdrix
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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?

    • 11 months ago
  11. Christos
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    1 ?

    • 11 months ago
  12. zepdrix
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    what? XD lol

    • 11 months ago
  13. Christos
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    uhm I think its the whole numerator? I am not sure :D

    • 11 months ago
  14. zepdrix
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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`.

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

    • 11 months ago
  16. zepdrix
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    The curve or the straight line? :D

    • 11 months ago
  17. Christos
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    numerator = denominator?

    • 11 months ago
  18. Christos
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    Eeeh i mean that high one not the straight one the other one grows faster

    • 11 months ago
  19. zepdrix
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    Yes good good, the curvy one grows faster.

    • 11 months ago
  20. zepdrix
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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.

    • 11 months ago
  21. zepdrix
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    That too confusing? :3 stick with the math rules?

    • 11 months ago
  22. Christos
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    same applies for x->-inf ? and what about the denominator having the highest power?

    • 11 months ago
  23. zepdrix
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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.

    • 11 months ago
  24. Christos
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    thanks

    • 11 months ago
  25. zepdrix
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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

    • 11 months ago
  26. Christos
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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.

    • 11 months ago
  27. reemii
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    Congrats!

    • 11 months ago
  28. UnkleRhaukus
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    Well done @Christos ! \(\LARGE\mathcal{CONGRATULATIONS!!}\)

    • 11 months ago
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