Need a little better explanation of limits please?

Question

The prof is Indian with a heavy accent so I'm having some trouble. OK I have the basic idea like for most functions you just replace x with whatever x is approaching. But how can you tell if there is a limit or the limit doesn't exist? He did something like for example:

lim

x->2+

lim

x->2-

The + and - are superscripts. How do you evaluate this stuff? :/

Answer 1

x->2+ means the limit as x approaches 2 from the right.  (i.e.: when it approaches 2 from values higher than 2.

x->2- means the limit as x approaches 2 from the left.  (i.e.: when it approaches 2 from values less than 2.

If you evaluate your function with both limits, and the two limits are not equal, then the limit does not exist when x approaches the value from both sides.

For example, consider the following function:

f(x) =

x when x>2

x+1 when x<=2

In this case, the limit as x->2+ is 2, while the limit as x->2- is 3.  Since the two are not equal, the limit of f(x) as x->2 does not exist.

Now consider f(x)=x

In this case, the limit as x->2+ and as x->2- are both equal to 2.  In this case, the limit of f(x) as x->2 does exist.

Answer 2

the + means the graph is coming from the right side

the - means the graph is coming from the left side

lim

x->2     is what the graph looks like it is going to as the x value gets close to 2

if

lim        and    lim         look like they are going to same place then

x->2+             x->2-

lim exists at where it look like it exists at

x->2

but if they dont look like they're going to the same place, then

lim  does not exist

x->2

Answer 3

You are correct, to a point. If the function is continuous and defined at the limit point, then plugging in the x-value will give you the limit. This implies that the limits approaching from both sides (the + and - you refer to) are equal.

As you get into limits further, you'll find problems like this:

f(x) = {1 for x ≥ 1

........ {0 for x < 1

What is lim{x->1} f(x)? The answer is undefined, since lim{x->1+} = 1 and lim{x->1-} = 0.

Also, consider lim{x->0} sin x / x. You might be tempted to say this is undefined, but you'll find later that it is, in fact, 1.