I want to find the limit of the piece-wise function as follows:
And here are the limit to find out
The expected answer is -1 and -1/3 respectively.
In the calculator with latest firmware (2020 01 21), it returns both limits to 0 which is not correct.
In the emulator (as in this picture), it provides error message for the first one, but provide the correct answer for the second one.
I verified this with TI Nspire, it provide the correct answer.
could you please help fix this issue.
Can you give the input as a commandline? Pictures are nice but can not be copied/pasted.
Definition of f1:
f1(x):=PIECEWISE(x>1,x^2,(x>0) AND (x<=1),x-1,x<=0,0)
The first limit:
limit(f1(x^2),x,0,-1)
The second limit:
limit((f1(x-1))/(x+2),x,1,1)
In the simulator if you replace x^2 with x*x you get -1 for the first limit:
limit(f1(x*x),x,0,-1) --> -1
-road
(06-02-2020 04:19 PM)roadrunner Wrote: [ -> ]In the simulator if you replace x^2 with x*x you get -1 for the first limit:
limit(f1(x*x),x,0,-1) --> -1
-road
Apparently the expression x^2<=0 simplifies to false (??).
It works with x*x because the calculator doesn't automatically simplify that, so the limit takes the expression x*x<=0.
Thanks, this is now fixed in the source code.
By the way, I'm not a big PRIME user, but let me say off-topic a thank you to Bernard for all his work on the CAS of multiple HP machines.
(06-02-2020 05:35 PM)victorvbc Wrote: [ -> ] (06-02-2020 04:19 PM)roadrunner Wrote: [ -> ]In the simulator if you replace x^2 with x*x you get -1 for the first limit:
limit(f1(x*x),x,0,-1) --> -1
-road
Apparently the expression x^2<=0 simplifies to false (??).
It works with x*x because the calculator doesn't automatically simplify that, so the limit takes the expression x*x<=0.
I don’t see it: how the 3rd condition could ever be true, even with x*x?
(06-02-2020 10:33 PM)pinkman Wrote: [ -> ]I don’t see it: how the 3rd condition could ever be true, even with x*x?
It's true if x=0
Thank you parisse for fixing this. Thank you all for your help.
(06-02-2020 10:50 PM)victorvbc Wrote: [ -> ] (06-02-2020 10:33 PM)pinkman Wrote: [ -> ]I don’t see it: how the 3rd condition could ever be true, even with x*x?
It's true if x=0
Yes I know that, but I don’t understand what you did understand:
Quote:Apparently the expression x^2<=0 simplifies to false (??).
It works with x*x because the calculator doesn't automatically simplify that, so the limit takes the expression x*x<=0.
What simplification could lead the algorithm to understand x*x is more <=0 than x^2?
It's not about the expression itself, just that the calculator doesn't evaluate x*x<=0 to false, like it does with x^2<=0. Only if you simplify it.
Here's a screenshot to illustrate my point:
[
attachment=8546]
So my theory is the limit program doesn't have any problems with handling the inequality x*x<=0, but it can't handle the direct false from x^2<=0. That's why it throws the "Unable to handle when condition is false error".
Edit: The correct simplification, of course, should just be x^2<=0. Because if x is zero (or some imaginary number), the resulting expression is true.
Thanks, I didn’t get it, now I do!