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anonymous

  • one year ago

Find the limit: lim as x approaches infinity 1-2x / sqrt(9x^2 +5)

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  1. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) like this?

  2. anonymous
    • one year ago
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    yes

  3. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle-\lim_{x \rightarrow ~\infty }\frac{(2x-1)}{\sqrt{9x^2+5}}}\) now apply LHS

  4. SolomonZelman
    • one year ago
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    I factored the top out of -1 and took the constant (the -1) out. as lim cf(x) = c lim f(x)

  5. anonymous
    • one year ago
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    no not yet

  6. SolomonZelman
    • one year ago
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    oh, then disregard what I said, since you haven't learned the L'Hospital's rule

  7. SolomonZelman
    • one year ago
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    are you asked to do this in a particular way, oor you can use any method? I would simply use a numerical approach

  8. SolomonZelman
    • one year ago
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    (to plug in very large number to see what the limit is going to approach)

  9. SolomonZelman
    • one year ago
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    or you can go about is intuitively

  10. anonymous
    • one year ago
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    we can use any method. I plugged in the equation into my calculator and everytime I'd plug in larger number it would get closer to -.66636 or some value around that.

  11. SolomonZelman
    • one year ago
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    intuitively: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) the top of the limit as x goes to very large numbers is just -2x, and the +1 there becomes even less significant the greater x value we choose. the bottom of the limit as x goes to very large numbers is just √(9x²) , becuase the +5 at some point (just like +1 on top) becomes insignificant. So you can write it as: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{1-2x}{\sqrt{9x^2+5}}}\) (and it will be an equivalent)

  12. SolomonZelman
    • one year ago
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    oh, I meant to say, to write it as \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}\)

  13. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{\sqrt{9x^2}}}\) simplify the square root \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\frac{-2x}{3x}}\) x's cancel, and you get?

  14. anonymous
    • one year ago
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    -2/3 ?

  15. SolomonZelman
    • one year ago
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    yup

  16. SolomonZelman
    • one year ago
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    this limit is -2/3

  17. anonymous
    • one year ago
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    ohhhh! makes so much sense. I had this problem on my exam..... I got it wrong

  18. anonymous
    • one year ago
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    thank you though for clarifying this that way I don't make the same mistake. Thank you.

  19. SolomonZelman
    • one year ago
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    Yw....

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