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burhan101

simple limit but i am blank

  • one year ago
  • one year ago

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  1. burhan101
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    \[\huge \lim_{x \rightarrow 5} \frac{ 1 }{ x-5 }\]

    • one year ago
  2. wio
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    Does it exist?

    • one year ago
  3. burhan101
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    yes

    • one year ago
  4. wio
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    Are you sure about that?

    • one year ago
  5. burhan101
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    not really

    • one year ago
  6. burhan101
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    i completely forgot how to solve these except for direct substitution

    • one year ago
  7. wio
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    Well, when you can't directly substitute, I recommend putting in really close values (e.g. 5.001 and 4.999).

    • one year ago
  8. wio
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    Make sure it exists first. Then you can use things like squeeze theorem / l'Hospital's rule. There is no sure way of doing limits, only a series of methods.

    • one year ago
  9. burhan101
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    okay thank you :)

    • one year ago
  10. burhan101
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    how would i solve limits that have fractions over fraction though ?

    • one year ago
  11. wio
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    Start by simplifying it into a single fraction.

    • one year ago
  12. burhan101
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    \[\large \lim x \rightarrow -3 \frac{ \frac{ 1 }{ x } +\frac{ 1 }{ 3 }}{x+3}\]

    • one year ago
  13. burhan101
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    ohh okay

    • one year ago
  14. wio
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    (a/b)/c = a/(bc) and a/(b/c) = (ac)/b

    • one year ago
  15. wio
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    Okay you need to add up the \(1/x\) and \(1/3\).

    • one year ago
  16. wio
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    a/b + c/d = (ad + bc)/(bd)

    • one year ago
  17. burhan101
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    \[\large \lim \rightarrow -3 \frac{ 3+x }{ 3x^2+9x }\]

    • one year ago
  18. wio
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    I don't think you're doing it correctly. \[ \frac{1}{x} +\frac{1}{3} = \frac{x+3}{3x} \]And then \[\Large \frac{\frac{x+3}{3x}}{x+3} = \frac{x+3}{3x(x+3)} = \frac{1}{3x} \]

    • one year ago
  19. wio
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    This function becomes continuous once you have manipulated it a bit.

    • one year ago
  20. wio
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    Continuous at \(-3\) at least.

    • one year ago
  21. burhan101
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    what does that mean ?

    • one year ago
  22. wio
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    What does continuous mean? It means: 1) \(f(a)\) is defined 2) \(\lim_{x \to a}f(x)\) exists and 3) \(\lim_{x \to a}f(x) = f(a) \) In general, it means you can just plug \(a\) into \(f(x)\) and get the answer to the limit.

    • one year ago
  23. burhan101
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    ohh thankyou !

    • one year ago
  24. Kainui
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    Ask yourself, what is a limit? Really. It's what happens as you approach a number from both sides. So to approach a number, you can plug in numbers close to it, like 4.999 is close to 5 on the left side while 5.00001 is close on the right side. Make sense? Plugging these in and seeing what you approach is really the essence of a limit and understanding that will let you solve any limit problem by simply graphing it when you get stumped.

    • one year ago
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