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#1 2006-12-02 07:46:24

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

limit question

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 hmm how does that work?

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#2 2006-12-02 10:02:37

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: limit question

well put it this way, if you just set h to 0, how would you deal with the division by h?


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#3 2006-12-02 14:18:36

Naresh
Member
Registered: 2006-09-29
Posts: 15

Re: limit question

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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#4 2006-12-02 14:25:09

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: limit question

but shouldn't the uppermost expression be exactly equal to the lower expression? and therefore theoretically yield the same answer?

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#5 2006-12-02 14:43:48

Naresh
Member
Registered: 2006-09-29
Posts: 15

Re: limit question

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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#6 2006-12-02 14:58:53

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: limit question

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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