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flexastexas

  • one year ago

Limits

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  1. flexastexas
    • one year ago
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  2. flexastexas
    • one year ago
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    I used my ti 84 to bring up the table and I was wondering

  3. flexastexas
    • one year ago
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    If 0 on the table says error what does this mean?

  4. flexastexas
    • one year ago
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    Does this mean there is no limit?

  5. flexastexas
    • one year ago
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  6. random231
    • one year ago
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    yeah that means the limit does not exists.

  7. IrishBoy123
    • one year ago
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    the limit is \(\frac{\sqrt{3}}{6}\) you should put \(x = \pm 0.00001\) in your calculator, experiment, but do not put x = 0 in there is a singularity at x = 0 which is why your claculator is crashing, it cannot divide by zero but this is a full on 2 sided limit

  8. flexastexas
    • one year ago
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    @IrishBoy123 I understand that the limit is what comes prior to the 0. In that case, The closest number I could get to 0 would be my limit?

  9. flexastexas
    • one year ago
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    So instead of using the 0 to determine if I have a limit or not, I use 0.0000001

  10. IrishBoy123
    • one year ago
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    yes, if you have not been taught how to find limits, eg L'Hopital's rule, then you put in a number that is very close to, in this case , zero you have to do it from both sides, so try , just as an example, -0.0000001 and +0.0000001 if these agree you have a proper 2-sided limit. in terms of actually solving, you can do this: \[\frac{\sqrt{3+x} - \sqrt{3}}{x}.\frac{\sqrt{3+x} + \sqrt{3}}{\sqrt{3+x} + \sqrt{3}}\] \[= \frac{3+x-3}{x(\sqrt{3+x} + \sqrt{3})}\] \[= \frac{x}{x(\sqrt{3+x} + \sqrt{3})}\] \[= \frac{1}{\sqrt{3+x} + \sqrt{3}}\] now set x to zero in that and you get the limit i suggested... or L'Hopital gives you straight out of the box \[\frac{1}{2 \sqrt{x+3}}\]

  11. IrishBoy123
    • one year ago
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    but if you have not learned this, just get very very close with the calculator. that works well!

  12. flexastexas
    • one year ago
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    Okay thanks for clarifying. From what I understand, if both sides of the table doesnt match, the limit does not exist.

  13. IrishBoy123
    • one year ago
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    https://gyazo.com/a249a4c50f358f942e4a7ef6a2519976 yes, the "ordinary" limit does not exist if they don't agree. [but you have right hand and left hand limits even though there is no ordinary limit]

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