15 Sep 08
Originally posted by FabianFnasPerhaps you should check the definition of limit. Remember that you're only in the neighbourhood of 0 and never exactly at 0. Since both functions are only 0 at x=0, then it's ok to do this operation within the limit operator.
lim 2x/3x you get a "0/0" expression on which you apply the l'Hôpital rule by your spinal cord.
But not knowing l'Hôpital rule you are in trouble again.
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15 Sep 08
1 edit
Well I thank you all...
Fabian, I see your point...but using lHopital is simply not an option at this point in the game for me.....perhaps i will gain the full understanding of this minor concept someday, assuming all good things come in time...for now however, x's divide out😵, but not without an underlying idea that things won't always be so straight forward...which is rewarding in its own way...
15 Sep 08
1 edit
Originally posted by joe shmoAt the earliest time of days, I got an advice from an older friend of mine. He said: "l'Hôpital is easy to learn, easy to use, skip the theory behind and just use it. Then you can do the limes calculations and check your result with l'Hôpital. You don't even admit to your teacher that you use it." I did so, and made some progress, I eliminated errors in the somewhat more complicated calculations. l'Hôpital is my friend.
Well I thank you all...
Fabian, I see your point...but using lHopital is simply not an option at this point in the game for me.....perhaps i will gain the full understanding of this minor concept someday, assuming all good things come in time...for now however, x's divide out😵, but not without an underlying idea that things won't always be so straight forward...which is rewarding in its own way...
16 Sep 08
Originally posted by FabianFnasx is not equal to 0 as you're calculating the limit of the function as x approaches 0. The function is undefined at x=0 which is the whole point of calculating the limit.
Well you cannot divide the numerator and denominator by x when x=0 just to get away the x's, because then you divide with zero. You just can't do it.
16 Sep 08
Not saying it is the case here, but a a function undefined for a particular value can have 2 different limits for the same undefined value, depending on the direction you approach it from.
Probably not the case here as it would be for something like tan(0), but worth considering if you are defining what a limit is..