Calculus Limits, Calculus help, TI-89 Titanium, Program

Uploaded by Tommynnnnn on 09.09.2012

This is a video on limits as it applies to calculus, generally calculus one, and let’s
get started. You have to press second alpha to get to my menu. You can scroll down from the menu many things
you want, all in alphabetical order, we’re going to go down to limits and press enter.
Here’s the choices in the limit program you have with regards to limits. You can compute
the limit, see the definition of a limit formula, complete a table of limits, and prove that
a limit is equal to the computed limit. You can also find delta given epsilon. We’re
going to do the first choice now finding delta given epsilon. You have to press alpha before
you enter anything in the boxes that come up in my programs. We’re going to try two
times x minus five equals one as x approaches three and the epsilon they give you, They will give you these things on the test problem they
give you, so epsilon equals point zero one. So I show you what you’ve entered. The limit
of two times x minus five equals one, as x approaches 3, while epsilon equals point zero
one. The next screen you have the choice of changing it or saying that it is ok. I say
it’s ok so I’ll press one for the choice. Here’s the first formula, absolute f of
x minus L is less than epsilon. I took the one over here and added it to the minus five
to get minus six equals zero. That’s an absolute value sign not a one, and that all
is less than point zero one. You’ll keep writing this stuff down on your paper exactly
as it shows. This is the way it done on youtube videos, or at least what I could find, and
delta equals point zero zero fine.
Now we can go back to choose main menu or new problem, we want to do a new problem so
I’ll choose one again and then go to number two and compute the limit. You can put anything
in you want within reason. Alpha again, anytime you are doing division you have to put parenthesis
in – in any problem – so we want to make sure we get the parenthesis in there. Let’s
go parenthisis x minus two closed parenthesis, divided by, parenthesis x squared minus x
minus two, closed parenthesis. As x goes to alpha two. Here’s the problem you’ve entered
and you want to compute the limit. You say it’s ok. Now at x equals two, f of x equals
zero and therefore doesn’t exist, so you use the following to find the limit. You add
point zero one to every x in the formula so you are a very little away from the limit
given and then compute it. The limit is point three, I give you other ways of showing that,
because some tests require other answer forms.
Let’s do another problem, let’s do a table of limits – number four – let’s clear
that out and do another one. Alpha closed parenthesis x minus closed parenthesis divided
by x squared minus four, as x approaches two, and here’s the table you’d mark down on
your paper.
Let’s do another problem, this is always fun. Prove that the limit is L – the limit
– this is done in always calculus one and throws every student off. Youre panicked because
they are talking about epsilon and delta and the definition of it, which sounds quite complicated,
and of course it is useless for the rest of calculus in your life. Let’s put a problem
in here. Alpha three times x plus five equals thirty-five as x approaches ten, You have
to press alpha again to enter the ten. and here’s what your problem is. Here’s the
formula again. F of x minus L is less than epsilon. You write this stuff down exactly
as shown on your paper. You factor it and then x minus ten is less than epsilon divided
by three. Epsilon becomes delta, and here’s the proof. If zero is less that absolute x
minus c – c is a constant – and is less that delta, then zero is less that absolute
x minus 10 is less than epsilon over three and therefore the original function is less
than epsilon. Put this on your paper and get one hundred percent on that problem. Everystepcalculus
dot com, check it out and check out the blogs also.