Question

Limits

Answer 1

I used my ti 84 to bring up the table and I was wondering

Answer 2

If 0 on the table says error what does this mean?

Answer 3

Does this mean there is no limit?

Answer 4

yeah that means the limit does not exists.

Answer 5

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

Answer 6

@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?

Answer 7

So instead of using the 0 to determine if I have a limit or not, I use 0.0000001

Answer 8

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
\[\frac{1}{2 \sqrt{x+3}}\]

Answer 9

but if you have not learned this, just get very very close with the calculator. that works well!

Answer 10

Okay thanks for clarifying. From what I understand, if both sides of the table doesnt match, the limit does not exist.

Answer 11

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]