Studying Mathematics Thread

[polymath257]

How about left-hand and right-hand limits:

Say, you have a function h(x) = sqrt(x). Since we're not dealing with imaginary numbers, the domain in this case would the set of all nonnegative numbers, so the least possible argument of the function would be 0.

What is the limit of h(x) as x approaches 0 from the right-hand side (the positive side)? Well, the function from the right to 0 is continuous and ends at the value of 0 at x = 0.

But what about the left-hand limit as x approaches 0 from the negative direction? Well, since we aren't dealing with negative x's in this case, then it seems the left-hand limit is undefined. Did I get this right? Or should I actually consider negative x's since they do lead to well-defined values (and, therefore, the limit would be 0 as well)? Or is the answer context-based?

You got this right. The function is only defined (well, for real values) for x>=0.

In general, if we are looking at the endpoints of an interval, we test continuity using the appropriate one-sided limit.

The next interesting expansion of this is to consider functions of more than one variable. So, we can look at things like the limit of  f(x,y)=x^2+y^3 as (x,y)->(1,2).

The only modification is that in the 'x' tolerance, delta, we consider the points (x,y) that are closer to (1,2) than delta. So,

For every epsilon>0, there is a delta>0 such that for every (x,y) with dist( (x,y), (1,2) )<delta we get |f(x)-L|<epsilon.

We use the Pythagorean theorem to compute the distance: dist( (x,y), (1,2) )=sqrt{ (x-1)^2 + (y-2)^2 }

After this, the definition for functions of any number of variables is easy. *grin*

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How about left-hand and right-hand limits:

Say, you have a function h(x) = sqrt(x). Since we're not dealing with imaginary numbers, the domain in this case would the set of all nonnegative numbers, so the least possible argument of the function would be 0.

What is the limit of h(x) as x approaches 0 from the right-hand side (the positive side)? Well, the function from the right to 0 is continuous and ends at the value of 0 at x = 0.

But what about the left-hand limit as x approaches 0 from the negative direction? Well, since we aren't dealing with negative x's in this case, then it seems the left-hand limit is undefined. Did I get this right? Or should I actually consider negative x's since they do lead to well-defined values (and, therefore, the limit would be 0 as well)? Or is the answer context-based?

Yes, this is what I meant by "from above" and "from below", that's the same as right hand and left hand limits. Yours is probably better terminology!

If a function is defined only for 0 upwards, then it seems to me you can't take a left hand limit at 0, because there are no values to consider.

Both terminologies are used. We also write lim x->0^+ for the limit from the right (above). Use a negative sign for the left (below).