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flexastexas
 one year ago
Limits
flexastexas
 one year ago
Limits

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flexastexas
 one year ago
Best ResponseYou've already chosen the best response.1I used my ti 84 to bring up the table and I was wondering

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.1If 0 on the table says error what does this mean?

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.1Does this mean there is no limit?

random231
 one year ago
Best ResponseYou've already chosen the best response.0yeah that means the limit does not exists.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1So instead of using the 0 to determine if I have a limit or not, I use 0.0000001

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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 2sided limit. in terms of actually solving, you can do this: \[\frac{\sqrt{3+x}  \sqrt{3}}{x}.\frac{\sqrt{3+x} + \sqrt{3}}{\sqrt{3+x} + \sqrt{3}}\] \[= \frac{3+x3}{x(\sqrt{3+x} + \sqrt{3})}\] \[= \frac{x}{x(\sqrt{3+x} + \sqrt{3})}\] \[= \frac{1}{\sqrt{3+x} + \sqrt{3}}\] 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
 one year ago
Best ResponseYou've already chosen the best response.1but if you have not learned this, just get very very close with the calculator. that works well!

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.1Okay thanks for clarifying. From what I understand, if both sides of the table doesnt match, the limit does not exist.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1https://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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