- #1

mathshead

can someone explain what the formal difinition of a limit ?

- Thread starter mathshead
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- #1

mathshead

can someone explain what the formal difinition of a limit ?

- #2

- 24

- 0

n -> [oo] An < +/- [oo]

- #3

- 54

- 1

for every number epsilon (usually, cant find the symbol) > 0, there is a corresponding number [sig] > 0, such that for all x;

0 < |x-a| < [sig] ==> |f(x) - L| < epsilon.

Is the formal definition.

for example (easy one);

To show that the limit of 5x - 3 as x tends to 1 is actually 2;

so a = 1, L = 2 (since this is what it does appear to converge to). Need

0 < |x-1| < [sig] for any epsilon > 0.

f(x) is within epsilon of L ie. |f(x) - 2| < epsilon. So to find [sig] from this,

|(5x-3) - 2| = |5x - 5| < epsilon

5|x - 1| < epsilon

|x - 1| < epsilon / 5.

so [sig] = epsilon / 5

and from 0 < |x - 1| < [sig] = epsilon / 5,

|(5x - 3) - 2| = 5 |x - 1| < 5 (epsilon / 5) = epsilon.

Which proves that L = 2.

Alternatively find a good text book

- #4

suffian

Given any allowable magnitude of error (formally epsilon) from a value (the limit), there exists a range near c (the value x is approaching) for which the function's outputs ( f(x) ) will deviate from the limit no more than the given magnitude of error (epsilon).

The key here is that if the limit for f(x) at a particular point c exists (and hence the previous statement holds), then we are stating that we can get f(x) as close to L as we want. I can make it within .001 or .000001, ... anything (because for each error I present to it, the limit existing garuntees that i can find an interval of values for x symmetrically about c such that f(x) will be that close to L).

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