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i was dabbling in some pre-calc and i was wondering why you can't just make the 'h' equal zero in this
when
instead of having to simplify to
before making the h a zero.
Do you always have to fully simplify an expression before you can apply the properties of limits how does that work?
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well put it this way, if you just set h to 0, how would you deal with the division by h?
The Beginning Of All Things To End.
The End Of All Things To Come.
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the answer would be
(x^2 + 2xh + h^2 - x^2)/h = (2xh+h^2 )/h = 2x + h.And yes You have to eliminate h from the bottom in this case as h tends to 0 then expression tends to infinite.Limit means the limiting value of the expression which is a finite not an infinite quantity unless the limit is an infinite limit which is not possible in this case.
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but shouldn't the uppermost expression be exactly equal to the lower expression? and therefore theoretically yield the same answer?
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No limit only defines the neighburhoood of that point but not the point itself.The point istself is the value of the function at that point but we want to know about the neighbourhood of that point so limit does not give the precise value but gives the limiting value.
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If f is continuous at a value a in its domain, then indeed we have that
However, the case we are considering is different. In the limit
let us consider the limit at a point x = a, so that
We can make the substitution h = x - a so that we have
(Note how we now have x approaching a, which makes sense because h = x - a approaches 0 as x approaches a.)
Then let us define the function
so that
Now, can we just say that
The answer is no. Why is this? We can see why if we note that g is not continuous at x = a, since
which is undefined. (To see why, assume 0/0 equals some number c. Then multiply each side by zero to get 0 = 0 * c. But this property holds for every number, so c = 0/0 cannot be defined.)
This would certaintly be a trouble then, since for ANY f(x), g(a) is undefined, and yet if we expand the limit and simplify, we can get an answer (for example with f(x) = x², the limit is 2x). Hopefully you can understand then why we cannot directly substitute h = 0 into the limit, since there does not exist continuity at h = 0.
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