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March 3, 2015

March 3, 2015

Posted by **glen** on Friday, July 6, 2007 at 3:32pm.

a) lim x->2 (x^2)/(x-1)

b) lim x->3 (x^2 - x - 12)/ (x - 3)

c) lim x->4 (x^2 - x - 12)/ (x - 3)

d) lim x->infinity (x^2 - 3x -2)/(x^3-5x)

a) just plug in x=2.

b) you cant plug in x=3, as that would give an indeterminate number (zero in denominator). So, factor the numerator, see if you then can divide out the denominator.

c) Plug in 4 for x.

d) at inf, the denominator cube will predomiate. YOu do this one by multipling numerator, and denominator by 1/x^3, then multiplying that into each term, then setting to inf.

The intuitive definition of a limit is as follows:

Lim x-->q f(x) is the value that f(x) will approach if you let x get closer and closer to q.

This limit does not have to be the same as f(q). If it is the same as f(q) then we say that the function f(x) is continuous at x = q.

The mathematical definition of the limit looks a bit complicated, but it is just a matter of translating the sentence "the value that f(x) will approach if you let x get closer and closer to q." into maths.

Suppose that lim x -->q f(x) = y

Then that means that if you let x get closer and closer to q then f(x) will get closer and closer to y. But how can we know if this is indeed the case?

The problem is that f(x) does not have to approach y in a smooth manner if you let x go closer and closer to q. So, it would not be correct to say that

|f(x) - y| should be a decreasing function in some small interval around x = q.

Instead we should turn things around and demand that for every interval around y, no matter how small, you should alway be able to find an interval around q that is mapped within that small interval around y by the function f(x).

Problem a):

lim x->2 (x^2)/(x-1) = 4

I.e. the limit in this case is just the value the function assumes at x = 2. This is intuitively clear, because you know that the function is continuous around x = 2 (the graph of the function doesn't show any jumps there).

If you want to prove that the limit is indeed 4, then you must use the definition of the limit. I.e. you must show that for every interval around 4 there exists an interval around x = 2 that is mapped by the function in that interval around 4.

We can do that as follows. We solve the equation:

(x^2)/(x-1) = y --->

x^2 = yx - y --->

x^2 - y x + y = 0 --->

x = y/2 ± sqrt[y^2/4 - y]

If we now choose an arbitrary interval around y = 4, e.g. from

4 - delta to 4 + delta then, using the above formula, you see that the interval

2 + delta/2 - sqrt[delta + delta^2] and

2 + delta/2 + sqrt[delta + delta^2]

is mapped into the interval for y around 4. This interval contains x = 2 for any delta, no matter how small you choose it.

- ineoflg orekuwah -
**ineoflg orekuwah**, Sunday, February 1, 2009 at 7:22pmqwsmrpl hsfpbozv nvdswz qwedpntgv oxmfbnzhu sdmpucbq kjvw

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