flexastexas
  • flexastexas
Limits
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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flexastexas
  • flexastexas
1 Attachment
flexastexas
  • flexastexas
I used my ti 84 to bring up the table and I was wondering
flexastexas
  • flexastexas
If 0 on the table says error what does this mean?

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flexastexas
  • flexastexas
Does this mean there is no limit?
flexastexas
  • flexastexas
random231
  • random231
yeah that means the limit does not exists.
IrishBoy123
  • IrishBoy123
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
flexastexas
  • flexastexas
@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?
flexastexas
  • flexastexas
So instead of using the 0 to determine if I have a limit or not, I use 0.0000001
IrishBoy123
  • IrishBoy123
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 \[\frac{1}{2 \sqrt{x+3}}\]
IrishBoy123
  • IrishBoy123
but if you have not learned this, just get very very close with the calculator. that works well!
flexastexas
  • flexastexas
Okay thanks for clarifying. From what I understand, if both sides of the table doesnt match, the limit does not exist.
IrishBoy123
  • IrishBoy123
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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