Limits involving ln(x)
We can use the rules of logarithms given above to derive the following
information about limits.
lim ln x = ∞,
x→∞
I
I
I
lim ln x = −∞.
x→0
We saw the last day that ln 2 > 1/2.
Using the rules of logarithms, we see that ln 2m = m ln 2 > m/2, for
any integer m.
Because ln x is an increasing function, we can make ln x as big as we
choose, by choosing x large enough, and thus we have
lim ln x = ∞.
x→∞
I
.
Similarly ln 21n = −n ln 2 < −n/2 and as x approaches 0 the values
of ln x approach −∞.
Example
Find the limit limx→∞ ln( x 21+1 ).
I As x → ∞, we have 21 → 0
x +1
I Letting u = 21 , we have
x +1
lim ln(
x→∞
1
) = lim ln(u) = −∞.
u→0
x2 + 1
Extending the antiderivative of 1/x
We can extend our antiderivative of 1/x( the natural logarithm function)
to a function with a larger domain by composing ln x with the absolute
value function |x|. . We have :
ln x
x >0
ln |x| =
ln(−x) x < 0
This is an even function with graph
We have ln|x| is also an antiderivative of 1/x with a larger domain than
ln(x).
Z
d
1
1
(ln |x|) =
and
dx = ln |x| + C
dx
x
x
Using Chain Rule for Differentiation
1
d
(ln |x|) =
dx
x
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I
d
g 0 (x)
(ln |g (x)|) =
dx
g (x)
and
Example 1: Differentiate ln | sin x|.
Using the chain rule, we have
1
d
d
ln | sin x| =
sin x
dx
(sin x) dx
I
=
cos x
sin x
Using Chain Rule for Differentiation : Example 2
Differentiate
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I
√
ln | 3 x − 1|.
We can simplify this to finding
√
ln | 3 x − 1| = ln |x − 1|1/3
d
dx
1
3
ln |x − 1| , since
d 1
1
1
d
1
ln |x − 1| =
(x − 1) =
dx 3
3 (x − 1) dx
3(x − 1)
Using Substitution
Reversing our rules of differentiation above, we get:
Z
Z 0
g (x)
1
dx = ln |x| + C and
dx = ln |g (x)| + C
x
g (x)
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I
R x
Example Find the integral
3−x 2 dx
Using substitution, we let u = 3 − x 2 .
du = −2x dx,
I
Z
I
=
x
dx =
3 − x2
x dx =
Z
du
,
−2
1
du
−2(u)
−1
−1
ln |u| + C =
ln |3 − x 2 | + C
2
2