16
1.2. THE NOTION OF LIMIT
First, as x → 3, it appears from the data (and the graph) that the function is
approaching approximately 0.866025. To be precise, we have to use the fact that
π
x →
π
3 ,
and thus we find that g(x) = sin(
π
x ) → sin(
π
3 ) as x → 3. The exact value of sin(
π
3 ) is
√
3
2 ,
which is approximately 0.8660254038. Thus, we see that
lim
x→3
g(x) =
√
3
2
.
As x → 0, we observe that
π
x does not behave in an elementary way. When x is
positive and approaching zero, we are dividing by smaller and smaller positive values,
and
π
x increases without bound. When x is negative and approaching zero,
π
x decreases
without bound. In this sense, as we get close to x = 0, the inputs to the sine function are
growing rapidly, and this leads to wild oscillations in the graph of g. It is an instructive
exercise to plot the function g(x) = sin
π
x
with a graphing utility and then zoom in on
x = 0. Doing so shows that the function never settles down to a single value near the
origin and suggests that g does not have a limit at x = 0.
How do we reconcile this with the righthand table above, which seems to suggest
that the limit of g as x approaches 0 may in fact be 0? Here we need to recognize that
the data misleads us because of the special nature of the sequence {0.1, 0.01, 0.001, . . .}:
when we evaluate g(10 −k ), we get g(10 −k ) = sin
π
10 −k
= sin(10 k π) = 0 for each positive
integer value of k. But if we take a different sequence of values approaching zero, say
{0.3, 0.03, 0.003, . . .}, then we find that
g(3 · 10
−k ) = sin
π
3 · 10 −k
= sin
10 k π
3
= −
√
3
2
≈ −0.866025.
That sequence of data would suggest that the value of the limit is
√
3
2 . Clearly the function
cannot have two different values for the limit, and this shows that g has no limit as x → 0.
An important lesson to take from Example 1.2 is that tables can be misleading when
determining the value of a limit. While a table of values is useful for investigating the
possible value of a limit, we should also use other tools to confirm the value, if we think
the table suggests the limit exists.
Activity 1.4.
Estimate the value of each of the following limits by constructing appropriate tables of
values. Then determine the exact value of the limit by using algebra to simplify the
function. Finally, plot each function on an appropriate interval to check your result
visually.
(a) lim
x→1
x 2 − 1
x − 1
1.2. THE NOTION OF LIMIT
First, as x → 3, it appears from the data (and the graph) that the function is
approaching approximately 0.866025. To be precise, we have to use the fact that
π
x →
π
3 ,
and thus we find that g(x) = sin(
π
x ) → sin(
π
3 ) as x → 3. The exact value of sin(
π
3 ) is
√
3
2 ,
which is approximately 0.8660254038. Thus, we see that
lim
x→3
g(x) =
√
3
2
.
As x → 0, we observe that
π
x does not behave in an elementary way. When x is
positive and approaching zero, we are dividing by smaller and smaller positive values,
and
π
x increases without bound. When x is negative and approaching zero,
π
x decreases
without bound. In this sense, as we get close to x = 0, the inputs to the sine function are
growing rapidly, and this leads to wild oscillations in the graph of g. It is an instructive
exercise to plot the function g(x) = sin
π
x
with a graphing utility and then zoom in on
x = 0. Doing so shows that the function never settles down to a single value near the
origin and suggests that g does not have a limit at x = 0.
How do we reconcile this with the righthand table above, which seems to suggest
that the limit of g as x approaches 0 may in fact be 0? Here we need to recognize that
the data misleads us because of the special nature of the sequence {0.1, 0.01, 0.001, . . .}:
when we evaluate g(10 −k ), we get g(10 −k ) = sin
π
10 −k
= sin(10 k π) = 0 for each positive
integer value of k. But if we take a different sequence of values approaching zero, say
{0.3, 0.03, 0.003, . . .}, then we find that
g(3 · 10
−k ) = sin
π
3 · 10 −k
= sin
10 k π
3
= −
√
3
2
≈ −0.866025.
That sequence of data would suggest that the value of the limit is
√
3
2 . Clearly the function
cannot have two different values for the limit, and this shows that g has no limit as x → 0.
An important lesson to take from Example 1.2 is that tables can be misleading when
determining the value of a limit. While a table of values is useful for investigating the
possible value of a limit, we should also use other tools to confirm the value, if we think
the table suggests the limit exists.
Activity 1.4.
Estimate the value of each of the following limits by constructing appropriate tables of
values. Then determine the exact value of the limit by using algebra to simplify the
function. Finally, plot each function on an appropriate interval to check your result
visually.
(a) lim
x→1
x 2 − 1
x − 1