Evaluation of limits

4.7 EVALUATION OF LIMITS
These can all be derived by straightforward application of Taylor’s theorem to
the expansion of a function about x = 0.
4.7 Evaluation of limits
The idea of the limit of a function f(x) as x approaches a value a is fairly intuitive,
though a strict definition exists and is stated below. In many cases the limit of
the function as x approaches a will be simply the value f(a), but sometimes this
is not so. Firstly, the function may be undefined at x = a, as, for example, when
f(x) =
sin x
,
x
which takes the value 0/0 at x = 0. However, the limit as x approaches zero
does exist and can be evaluated as unity using l’Hôpital’s rule below. Another
possibility is that even if f(x) is defined at x = a its value may not be equal to the
limiting value limx→a f(x). This can occur for a discontinuous function at a point
of discontinuity. The strict definition of a limit is that if limx→a f(x) = l then
for any number however small, it must be possible to find a number η such that
|f(x)−l| < whenever |x−a| < η. In other words, as x becomes arbitrarily close to
a, f(x) becomes arbitrarily close to its limit, l. To remove any ambiguity, it should
be stated that, in general, the number η will depend on both and the form of f(x).
The following observations are often useful in finding the limit of a function.
(i) A limit may be ±∞. For example as x → 0, 1/x2 → ∞.
(ii) A limit may be approached from below or above and the value may be
different in each case. For example consider the function f(x) = tan x. As x tends
to π/2 from below f(x) → ∞, but if the limit is approached from above then
f(x) → −∞. Another way of writing this is
lim tan x = ∞,
lim tan x = −∞.
x→ π2 −
x→ π2 +
(iii) It may ease the evaluation of limits if the function under consideration is
split into a sum, product or quotient. Provided that in each case a limit exists, the
rules for evaluating such limits are as follows.
(a) lim {f(x) + g(x)} = lim f(x) + lim g(x).
x→a
x→a
x→a
(b) lim {f(x)g(x)} = lim f(x) lim g(x).
x→a
x→a
x→a
limx→a f(x)
f(x)
=
, provided that
(c) lim
x→a g(x)
limx→a g(x)
the numerator and denominator are
not both equal to zero or infinity.
Examples of cases (a)–(c) are discussed below.
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SERIES AND LIMITS
Evaluate the limits
lim(x2 + 2x3 ),
lim(x cos x),
x→1
lim
x→0
x→π/2
sin x
.
x
Using (a) above,
lim(x2 + 2x3 ) = lim x2 + lim 2x3 = 3.
x→1
x→1
x→1
Using (b),
lim(x cos x) = lim x lim cos x = 0 × 1 = 0.
x→0
x→0
x→0
Using (c),
lim
x→π/2
limx→π/2 sin x
1
2
sin x
=
=
= .
x
limx→π/2 x
π/2
π
(iv) Limits of functions of x that contain exponents that themselves depend on
x can often be found by taking logarithms.
Evaluate the limit
lim
x→∞
Let us define
1−
a2
x2
y=
1−
x2
a2
x2
.
x2
and consider the logarithm of the required limit, i.e.
a2
.
lim ln y = lim x2 ln 1 − 2
x→∞
x→∞
x
Using the Maclaurin series for ln(1 + x) given in subsection 4.6.3, we can expand the
logarithm as a series and obtain
2
a
a4
lim ln y = lim x2 − 2 − 4 + · · ·
= −a2 .
x→∞
x→∞
x
2x
Therefore, since limx→∞ ln y = −a2 it follows that limx→∞ y = exp(−a2 ). (v) L’Hôpital’s rule may be used; it is an extension of (iii)(c) above. In cases
where both numerator and denominator are zero or both are infinite, further
consideration of the limit must follow. Let us first consider limx→a f(x)/g(x),
where f(a) = g(a) = 0. Expanding the numerator and denominator as Taylor
series we obtain
f(a) + (x − a)f (a) + [(x − a)2 /2!]f (a) + · · ·
f(x)
=
.
g(x)
g(a) + (x − a)g (a) + [(x − a)2 /2!]g (a) + · · ·
However, f(a) = g(a) = 0 so
f (a) + [(x − a)/2!]f (a) + · · ·
f(x)
= .
g(x)
g (a) + [(x − a)/2!]g (a) + · · ·
142
4.7 EVALUATION OF LIMITS
Therefore we find
f (a)
f(x)
= ,
x→a g(x)
g (a)
lim
provided f (a) and g (a) are not themselves both equal to zero. If, however,
f (a) and g (a) are both zero then the same process can be applied to the ratio
f (x)/g (x) to yield
f (a)
f(x)
= ,
x→a g(x)
g (a)
lim
provided that at least one of f (a) and g (a) is non-zero. If the original limit does
exist then it can be found by repeating the process as many times as is necessary
for the ratio of corresponding nth derivatives not to be of the indeterminate form
0/0, i.e.
f (n) (a)
f(x)
= (n) .
x→a g(x)
g (a)
lim
Evaluate the limit
lim
x→0
sin x
.
x
We first note that if x = 0, both numerator and denominator are zero. Thus we apply
l’Hôpital’s rule: differentiating, we obtain
lim(sin x/x) = lim(cos x/1) = 1. x→0
x→0
So far we have only considered the case where f(a) = g(a) = 0. For the case
where f(a) = g(a) = ∞ we may still apply l’Hôpital’s rule by writing
lim
x→a
f(x)
1/g(x)
= lim
,
g(x) x→a 1/f(x)
which is now of the form 0/0 at x = a. Note also that l’Hôpital’s rule is still
valid for finding limits as x → ∞, i.e. when a = ∞. This is easily shown by letting
y = 1/x as follows:
lim
x→∞
f(x)
f(1/y)
= lim
g(x) y→0 g(1/y)
−f (1/y)/y 2
= lim
y→0 −g (1/y)/y 2
f (1/y)
= lim y→0 g (1/y)
f (x)
= lim .
x→∞ g (x)
143