Question about limits in calculus?

Question

I just started taking calculus and the other day my teacher drew a graph of a curve on the board that intersected the x axis. He asked the class what the limit would be as x approaches 0. I was alittle confused by this because I don't understand how there can be a limit as x approaches 0 when the line itself actually goes through 0 and is therefore not approaching 0. Can someone please explain? Thanks.

Sorry, I meant to say that the graph intersected the y axis.

Answer 1

Did the teacher just draw a graph of y = f(x) as a simple curve that intersected the y axis?

If so, you're right to think this is almost "too simple". After all,

f(x) approaches f(0) as x approaches 0

or

f(x) -> f(0) as x->0

The curve goes directly through the point (0, f(0)) when you approach it from either side. Why bother talking about limits?

But, he was just giving you a very simple case before going on to something more complicated-- and more interesting. Suppose, instead, he'd drawn the graph of a function like

f(x) = x * sin(1/x)

which isn't defined when x=0. The graph wiggles up and down, with shorter-and-shorter period, as x gets close to 0, but it's always between

y = x

and

y = -x

because sin(x) is always between -1 and 1.

Now, what's the limit of f(x) as x approaches 0?

x*sin(1/x) -> 0 as x->0

even though the function isn't defined at x=0.

Answer 2

Well, in order to intersect the y-axis at x = 0, it would have to approach some value. In the case of what the teacher drew, it actually did intersect, but it doesn't have to.
Ex: If the teachers graph was y = 2x + 3, it would intersect the y-axis at (0,3). If you follow the line towards (0,3) from either the left or the right sides, you can see that the y values are approaching 3. Eventually, once you get to the y-axis, you are at 3.
Some graphs, however, will approach some value, but never acutally get there.
Ex:
y = 2x + 3, if x<0 or x>0
& y = 6, if x = 0
From this you can see the graph of y = 2x + 3 would have a hole in it at x = 0, because at x = 0, y = 6.
Therefore the limits approaching 0 from the left (-x) or the right (x) would appoach 3, but at 0 it would actually be 6

Answer 3

Maybe he was demonstrating how a limit can differ from 0- (from the negative x-axis) vs. 0+ (at the positive x-axis) if the function is not continuous at x=0

Answer 4

The limit as x approaches 0 will be the y intercept, not the x intercept.