Sorry for posting this up late Ms.Hwang I thought i had posted it up earlier...(:
oooopppps.(:
At first the sine functions were confusing
but when Ms. Hwang(you) explained to me it was more clear, i thought the limit when x aproaches 0 was the same as when the limit aproaches infinity.
Now i know the difference(:
I wont forget(: lol!
The piecewise functions are a little more challenging because there is more than one equation in the problem, making the problem that much harder.
Those problems are pretty hard and i would apprectiate more help on them(:
Problems i also get stuck with is like #5 on p.95
when the limit aproaches 0.... and it is the denominator.
that confuses me.. with the extended functions....?
Gosh../:
I need help(:
and would appreciate some(:
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for #5,make the top part of the fraction have a common denominator of 2(2+x). First solve the top and you get -x/4+2x, then divide it by the x that was on the bottom of the fraction (you really have to multiply by 1/x instead). then you get -1/4+2x. when you plug in 0 for x (since the original equation said to find the limit as x approaches 0), you get -1/4.
ReplyDeleteI hope this helped with you #5 problem (:
Piecewise functions can definitely get hard. Most of the time with limits though, we're normally only dealing with one side, either from the + side or the - side of that boundary. If so, then you only have to choose the equation that is relevant.
ReplyDeleteIf they ask you for the limit as x-->c without a + or - sign after it, that means that the limit has to be the same from BOTH sides together.