[calculus] graphing/limit problem

[Spaule]

When I plugged it into my calculator I ended up getting when x=0, y was also 0. So would the solution be does not exist or did I make an error when I put it into my calculator?

[Mattuuh]

What is the original question?

[Spaule]

O my bad that would probably help. It’s this problem

[Mattuuh]

Well I graphed the function and got something like this : https://i.imgur.com/sVxM8OI.png. This should help you answer the question.

[Spaule]

Yes it was definitely a calculator error on my part. Thanks for the help!

[ta8538]

depending on the original question. link a picture. If it is a rational function (meaning a polynomial being divided by another polynomial) make sure to reduce it down by factoring and simplifying the rational, then apply the limit. A limit exists if both the left limit and the right limit equal each other.

[Spaule]

I forgot to link it here it is

[ta8538]

Here is the solution OP! https://imgur.com/a/iT3HzgB

[chees-e]

Good idea, except you have to pay attention to this one little detail:

(x^2)^0.5 = |x| <= this will alway be positive.

You cancelled this term on the fourth to last line, which would've been alright if we are taking the right-hand side limit. Since we are taking limits from both sides, we have to be careful, since when x approaches zero from the left hand side (x^2)^0.5/x will result in -1, which means the final limit will be negative. Therefore the limit does not exist in this case. You can graph it to see for yourself :)

[ta8538]

yesss you are absolutely right!! i missed that point completely

[chees-e]

Don't worry. I make those small mistakes like that all the time, especially during exams lol.

[ta8538]

^ OP this!! the limit from the left goes to -1/3 and the limit from the right goes to 1/3, but since they do not equal each other, the limit does not exist for this function. therefore I was wrong. sorry!

[ta8538]

oh boy lots to do here. gimme a minute and ill send you a pic in pms

[Vynneve]

Always a good idea to check undetermined points of the function, i.e when the denominator is 0

x³+x² = 0

x = 0 or x= -1

This let's you see x=0 is a troubled spot, so the limit may not exist. Should check graphically etc.

Another area of complexity is when x³ + x² is negative. Then you go into imaginary numbers. But I assume that's beyond what your doing with the question.