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Explain the statement (Help) (1 Viewer)
- Thread starter ElizaBan6
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sida1049
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in the first photo, if u = x, then u - x = 0. This would have made the transition from the third line to the fourth be impossible, because it involves a division by (u - x) [which cancels out (u - x) from top and bottom], and if u - x = 0, then you wouldn't be able to do that (because you cannot divide by zero).
Similarly, in the second photo, they also divide by h (in order to cancel out h from top and bottom). And if h = 0, then that would not be possible.
So those "since..." statements justify the division (and cancellation) taken from the third line to the fourth.
Similarly, in the second photo, they also divide by h (in order to cancel out h from top and bottom). And if h = 0, then that would not be possible.
So those "since..." statements justify the division (and cancellation) taken from the third line to the fourth.
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Actually in Maths Ext 1 & 2, the concept of "limits" is treated rather superficially. It is true that if x = u, the denominator would = 0, and that would imply division by 0. So it says x =/= 0. But your textbooks don't explain why this is so. The answer is, in considering the process of finding the limit as x tend to u, we do not need to consider the case of x = u. Then you may well ask: why so? Well it is simply because you do not need to consider this case in the definition of limits; you don't care about the point x=u. To understand this, you will need to delve into the actual definition of the limit. That is outside the scope of the HSC syllabus.
So, in MX1 most students will say that "the limit of (x^2 -x -6)/(x-3) as x tends to 3" = 3+2 = 5, saying you cancel the common factor (x-3) without fully understanding the process, since this require x-3 to be nonzero.
So, in MX1 most students will say that "the limit of (x^2 -x -6)/(x-3) as x tends to 3" = 3+2 = 5, saying you cancel the common factor (x-3) without fully understanding the process, since this require x-3 to be nonzero.
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Seems legit. Ha ha...
So just to confirm, is this for all cases?
For example, h is not equal to 0 is used every single time? for the formula of f(x+h)....
And for the f(u)..... case, do we write u is not equal to x every time?
(Basically, it is learnt that you just do that and it just comes off as part of the rote learnt equation)
So just to confirm, is this for all cases?
For example, h is not equal to 0 is used every single time? for the formula of f(x+h)....
And for the f(u)..... case, do we write u is not equal to x every time?
(Basically, it is learnt that you just do that and it just comes off as part of the rote learnt equation)
]You probably would only need to write it in a step where you did something like cancellation. The value of a limit is unaffected by the function's value at the point we are taking the limit, and only affected by the behaviour of the function 'near' that point, which is the reason why we can do that (say h is not equal to 0 etc.). This follows from the definition of limits (which isn't in the HSC syllabus).
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