Indeterminate
Forms
In this
section, we shall discuss various “indeterminate forms” and methods of
evaluating the limits when we come across them.
While
computing the limits
lim x→α R(x)
of
certain functions R( x) , we may come across the following
situations like,
We say
that they have the form of a number. But values cannot be assigned to them in a
way that is consistent with the usual rules of addition and mutiplication of
numbers. We call these expressions indeterminate forms. Although they are not
numbers, these indeterminate forms play a useful role in the limiting behaviour
of a function.
John
(Johann) Bernoulli discovered a rule using derivatives to compute the limits of
fractions whose numerators and denominators both approach zero or ∞.
The rule is known today as l’Hôpital’s Rule (pronounced as Lho pi tal Rule),
named after Guillaume de l’Hospital’s, a French nobleman who wrote the earliest
introductory differential calculus text, where the rule first appeared in
print.
Suppose f ( x)
and g ( x) are differentiable functions and g(x′) ≠ 0 with
Example 7.33
Solution
If we
put directly x =
1 we observe that the given function is in an indeterminate form 0/0. As the numerator
and the denominator functions are polynomials of degree 2 they both are
differentiable. Hence, by an application of the l’Hôpital Rule, we get
Note
that this limit may also be evaluated through the factorization of the
numerator and denominator as
Example 7.34
Solution
If we
put directly x =a we observe that the
given function is in an indeterminate form 0/0 . As the numerator and the
denominator functions are polynomials they both are differentiable.
Hence by
an application of the l’Hôpital Rule we get,
Example 7.35
Solution
If we
directly substitute x =
0 we get an indeterminate form 0/0 and
hence we apply the l’Hôpital’s rule to evaluate the limit as,
The next
example tells that the limit does not exist.
Example 7.36
Solution
If we
directly substitute x =
0 we get an indeterminate form 0/0 and hence we apply the l’Hôpital’s rule to
evaluate the limit as,
As the
left limit and the right limit are not the same we conclude that the limit does
not exist.
Remark
One may be tempted to use the l’Hôpital’s rule once again in
which is
not true because it was not an indeterminate form.
Example 7.37
Solution
As this
is an indeterminate form (0/0), using the l’Hôpital’s Rule
Now
using the example 7.35, we have
Example 7.38
Solution
This is
an indeterminate form ∞/∞ and hence we use the l’Hôpital’s Rule to evaluate the
limit.
Example 7.39
Solution
This is
an indeterminate of the form ∞ − ∞ . To evaluate this limit we first
simplify and bring it in the form (0/0) and applying the l’Hôpital Rule, we get
Example 7.40
Evaluate :
Solution
This is
an indeterminate of the form (0 × ∞) . To evaluate this limit, we first
simplify and bring it to the form (∞/∞) and apply l’Hôpital Rule. Thus, we get
Example 7.41
Solution
This is
an indeterminate of the form (∞/∞) . To evaluate this limit, we apply
l’Hôpital Rule.
Example 7.42
Solution
This is
an indeterminate of the form (∞/∞).
To
evaluate this limit, we apply l’Hôpital Rule m times.
Thus, we have
limx→∞
ex/xm = limx→∞ ex/m!
= ∞ .
In order
to evaluate the indeterminate forms like this, we shall first state the theorem
on the limit of a composite function.
Theorem 7.6
Let limx→α g ( x) exist and let it be L and let f (x) be a continuous function at x = L . Then,
(1) Let A = limx→α g(x)
. Then taking logarithm, with the assumption that A > 0 to ensure the continuity of the
logarithm function, we get log A =
limx→α log(g(x)) . Therefore using the above theorem
with f (x) = log x we have the limit
(2) We
have the limit lim x→α log(g
( x)) into either (0/0) or (∞/∞) form evaluate it using l’Hôpital Rule.
(3) Let
that evaluated limit be say α
. Then the required limit is eα .
Example 7.43
Using
the l’Hôpital Rule, prove that limx→α (1+
x )1/x = e
.
Solution
This is
an indeterminate of the form 1∞ . Let g ( x) =
(1+
x)1/x. Taking the logarithm, we get
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