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AP.CALC:

LIM‑1 (EU)

, LIM‑1.E (LO)

, LIM‑1.E.1 (EK)

Let's say that f of x is equal
to x squared plus x minus 6 over x minus 2. And we're curious about
what the limit of f of x, as x approaches
2, is equal to. Now the first attempt
that you might want to do right when you
see something like this, is just see what happens
what is f of 2. Now this won't always be the
limit, even if it's defined, but it's a good
place to start, just to see if it's something
reasonable could pop out. So looking at it this way,
if we just evaluate f of 2, on our numerator, we get
2 squared plus 2 minus 6. So it's going to be 4 plus
2, which is 6, minus 6, so you're going to
get 0 in the numerator and you're going to get
0 in the denominator. So we don't have, the
function is not defined, so not defined at x is equal 2. f not defined. So there's no
simple thing there. Even if this did evaluate, if
it was a continuous function, then the limit would be
whatever the function is, but that doesn't
necessarily mean the case. But we see very clearly the
function is not defined here. So let's see if we
can simplify this and also try to
graph it in some way. So one thing that might
have jumped out at your head is you might want to factor
this expression on top. So if we want to
rewrite this, we can rewrite the top expression. And this just goes back
to your algebra one, two numbers whose
product is negative 6, whose sum is positive 3,
well that could be positive 3 and negative 2. So this could be
x plus 3 times x minus 2, all of
that over x minus 2. So as long as x
does not equal 2, these two things
will cancel out. So we could say this is equal
to x plus 3 for all X's except for x is equal to 2. So that's another
way of looking at it. Another way we could
rewrite our f of x, we'll do it in blue, just
to change the colors, we could rewrite f of x, this
is the exact same function, f of x is equal to x plus
3 when x does not equal 2. And we could even say it's
undefined when x is equal to 2. So given this definition,
it becomes much clearer to us of how we can
actually graph f of x. So let's try to do it. So that is, that is
not anywhere near being a straight line,
that is much better. So let's call this the y-axis
call it y equals f of x. And then let's,
over here, let me make a horizontal line,
that is my x-axis. So defined this way, f of
x is equal to x plus 3. So if this is 1, 2, 3, we
have a y-intercept at 3 and then the slope is 1. And it's defined for all X's
except for x is equal to 2. So this is x is equal
to 1, x is equal to 2. So when x is equal
to 2 it is undefined. So let me make
sure I can, so it's undefined right over there. So this is what f
of x looks like. Now given this, let's try
to answer our question. What is the limit of f
of x as x approaches 2. Well, we can look
at this graphically. As x approaches 2 from
lower values in 2, so this right over here is x is
equal to 2, if we get to maybe, let's say this is 1.7,
we see that our f of x is right over there. If we get to 1.9, our f
of x is right over there. So it seems to be approaching
this value right over there. Similarly, as we approach 2
from values greater than it, if we're at, I don't know,
this could be like 2.5, 2.5 our f of x is
right over there. If we get even closer to 2,
our f of x is right over there. And once again, we look like
we are approaching this value. Or another way of
thinking about it, if we ride this line from
the positive direction, we seem to be approaching
this value for f of x, if we write this line
from the negative direction, from values less than 2,
we seem to be approaching this value right over here. And this is essentially
the value of x plus 3 if we set x is equal to 2. So this is essentially going to
be, this value right over here, is equal to 5. If we just look at it
visually, if we just graphed a line with slope 1
with the y-intercept of 3, this value right over here is 5. Now we could also try to
do this it numerically, so let's try to do that. So if this is our
function definition, completely identical to
our original definition, then we just try values as x
gets closer and closer to 2. So let's try values less than 2. So 1.9999, and this
is almost obvious. 1.9999 plus 3, well, that gets
you pretty darn close to 5. If I put even more 9s
here, get even closer to 2, we'd get even closer to 5 here. If we approach 2 from
the positive direction, and then, we once again, we're
getting closer and closer to 5 from the
positive direction. If we were even closer to
2, we'd be even closer to 5. So whether we look
at it numerically, or we look at it graphically,
it looks pretty clear that the limit here is
going to be equal to five.

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