1. i'm currently reading on my own about limits, and the only "x approaches" exposure i've had before is "as x approaches infinity" in connection with Euler's Law. Obviously, x approaches, but never actually reaches, infinity. However, if you have, for example, "x--->2", does x actually reach 2, or does it just grow ever-closer to it without ever actually reaching it?  2.

3. It may help to give the actual definition and to unpack it.

Definition: A function f(x) from R to R is said to approach L as x approaches t iff for every epsilon > 0 there exists a delta > 0 such that if |x-t| < delta then |f(x) - L| < epsilon.

Now the trick here is that delta is greater then zero, so x is never equal to t and so x never reaches t (or 2 in your question).  4. that's the answer i was looking for, thanks  5. well, x can be equal to t because delta is just an upper bound on |x-t|.  6. Oops - your right, here is the corrected definition

Definition: A function f(x) from R to R is said to approach L as x approaches t iff for every epsilon > 0 there exists a delta > 0 such that if 0 < |x-t| < delta then |f(x) - L| < epsilon.  7. consider this case

limit (x-2)/(x-4)
x-->2

in this case you can assume that x takes the value 2 as there is no discontinuity or abberation here

the answer for the above limit is 0

but in this case

lim (x-2)/(x-2) the denominator tends to 0 as x keeps moving towards 2
x-->2

when x=2, the expression becomes 0/0 which is indeterminate. so we assume that x only approches 2 and see what becomes of the expression. as x tends to 2 the value of the function is the value to which the expression points to

in this case the expression tends to 1 as x tends to two and the expression is not defined when x equals two  8. Originally Posted by hypercube
consider this case

limit (x-2)/(x-4)
x-->2

in this case you can assume that x takes the value 2 as there is no discontinuity or abberation here

the answer for the above limit is 0

but in this case

lim (x-2)/(x-2) the denominator tends to 0 as x keeps moving towards 2
x-->2

when x=2, the expression becomes 0/0 which is indeterminate. so we assume that x only approches 2 and see what becomes of the expression. as x tends to 2 the value of the function is the value to which the expression points to

in this case the expression tends to 1 as x tends to two and the expression is not defined when x equals two
makes sense, thanks for taking the time to reply.

gotta love those hypercubes...   9. any time chemy any time :-D  10. iam sravan......just recently joined this forum

ur question totally depends on the function ur dealing with

case 1:

if the function is discontinuous at X=2 then X actually will be in the smallest neigbourhood around 2 then u have the L'Hospitals Rule for caluculating the value of function at X=2

case 2:

if the function is continuous at X=2 the u can just substitute the value of X in the function and get the value of the function directly.........here X actually approaches 2  Posting Permissions
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