TUESDAY'S MATH CLASS

Someone forgot to scribe for Tuesday's math class and she's making up for it by scribing for Tuesday and Thursday. Sorry. Well, on Tuesday we mainly focused on limits in a symbolic approach. The first question was....

Someone forgot to scribe for Tuesday's math class and she's making up for it by scribing for Tuesday and Thursday. Sorry. Well, on Tuesday we mainly focused on limits in a symbolic approach. The first question was....

- Factor both the numerator and denominator. If you don't factor the numerator and denominator and go straight to substituting in the value 2 for x, you'll end up with zero in the denominator making the whole thing undefined.
- Reduce
- After reducing the same terms in the numerator and denominator you substitute the value two in everywhere there is an x.
- Voila, you end up with the answer.

- Rationalize the numerator because if you do then you can reduce 25 - x in the numerator and denominator.
- After reducing, you're left with 1 over 5 + √x. You can now substitute the value 25 for x because it's in its most reduced form.
- Now simplify, the √25 is 5. What is left is 1 over 10.
- The numerator can be factored so that something can be reduced from the bottom.
- After reducing to the simplest form you now can substitute the value nine for where there's an x.
- The end result is negative 6.

- The whole thing is undefined because in the end your gonna have to substitute zero in for x and 3 over 0 is undefined. Any number over zero is undefined.

- That would have been marked wrong if it was on a test or exam. Why?

Well, in the end since it's just the notation and you've solved for it, you don't need the lim thing.

- For the rest of the class, we briefly talked about horizontal and vertical asymptotes. Homework was posted in the slides.

## No comments:

## Post a Comment