L'Hôpital's Rule
Occasionally, one comes across a limit which results in
or
, which are called indeterminate limits. However, it is still possible to solve these in many cases due to L'Hôpital's rule. This rule also is vital in explaining how a number of other limits can be derived.
All of the following expressions are indeterminate forms.
![{\displaystyle {\frac {0}{0}},{\frac {\pm \infty }{\pm \infty }},\infty -\infty ,0\cdot \infty ,0^{0},\infty ^{0},1^{\infty }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1ef48389a9d2b3b0876ae0760517fca88307d9b)
These expressions are called indeterminate because you cannot determine their exact value in the indeterminate form. Depending on the situation, each indeterminate form could evaluate to a variety of values.
Theorem
If
is indeterminate of type
or
,
then
In other words, if the limit of the function is indeterminate, the limit equals the derivative of the top over the derivative of the bottom. If that is indeterminate, L'Hôpital's rule can be used again until the limit isn't
or
.
Note:
can approach a finite value
,
or
.
Proof of the
case
Suppose that for real functions
,
and that
exists. Thus
and
exist in an interval
around
, but maybe not at
itself. This implies that both
are differentiable (and thus continuous) everywhere in
except perhaps at
. Thus, for any
, in any interval
or
,
are continuous and differentiable, with the possible exception of
. Define
![{\displaystyle {\begin{array}{l}F(x)={\begin{cases}f(x)&x\neq a\\\lim \limits _{x\to a}f(x)&x=a\end{cases}}\\G(x)={\begin{cases}g(x)&x\neq a\\\lim \limits _{x\to a}g(x)&x=a\end{cases}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d5a6ed278a1eb207febb5396748e73234e5b449)
Note that
,
and that
are continuous in any interval
or
and differentiable in any interval
or
when
.
Cauchy's Mean Value Theorem tells us that
for some
or
. Since
, we have
for
.
Note that since
or
, by the squeeze theorem
![{\displaystyle \lim _{x\to a}x=a\quad \Rightarrow \quad \lim _{x\to a}c=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cc8ea5c801baede8df9ddcc9a3bbfdf17b9517e)
This implies
![{\displaystyle \lim _{x\to a}{\frac {F'(c)}{G'(c)}}=\lim _{x\to a}{\frac {F'(x)}{G'(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58057b8f05e87c39b9dab56f452f1e8fc6faba8d)
So taking the limit as
of the last equation gives
which is equivalent to
.
Examples
Example 1
Find
Since plugging in 0 for x results in
, use L'Hôpital's rule to take the derivative of the top and bottom, giving:
![{\displaystyle \lim _{x\to 0}{\frac {{\frac {d}{dx}}\left(\sin(x)\right)}{{\frac {d}{dx}}(x)}}=\lim _{x\to 0}{\frac {\cos(x)}{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4531f26308c12ff3bfbae9ea627f789848b77a21)
Plugging in 0 for x gives 1 here.
Note that it is logically incorrect to prove this limit by using L'Hôpital's rule, as the same limit is required to prove that the derivative of the sine function exists: it would be a form of begging the question
Example 2
Find
First, you need to rewrite the function into an indeterminate limit fraction:
![{\displaystyle \lim _{x\to 0}{\frac {x}{\tan(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aab53b8d2b119e0c29459b680d3dda90d55c119)
Now it's indeterminate. Take the derivative of the top and bottom:
![{\displaystyle \lim _{x\to 0}{\frac {1}{\sec ^{2}(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/575528b22c25b9b0997d41c5df7c69470ed0c53a)
Plugging in 0 for x once again gives 1.
Example 3
Find
This time, plugging in
for x gives you
. You know the drill:
![{\displaystyle \lim _{x\to \infty }{\frac {4}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/967ea7452e13a97a7464edf191313c71caeb2ce9)
This time, though, there is no x term left!
is the answer.
Example 4
Sometimes, forms exist where it is not intuitively obvious how to solve them. One might think the value
However, as was noted in the definition of an indeterminate form, this isn't possible to evaluate using the rules learned before now, and we need to use L'Hôpital's rule to solve.
Find
Plugging the value of x into the limit yields
(indeterminate form).
Let
|
|
|
|
|
(indeterminate form)
|
We now apply L'Hôpital's rule by taking the derivative of the top and bottom with respect to
.
![{\displaystyle {\frac {d}{dx}}\left[\ln \left(1+{\frac {1}{x}}\right)\right]={\frac {x}{x+1}}\cdot {\frac {-1}{x^{2}}}=-{\frac {1}{x(x+1)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef39a8560ad1bae995f9459351d415c06be0b419)
![{\displaystyle {\frac {d}{dx}}\left({\frac {1}{x}}\right)=-{\frac {1}{x^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/787be0ba3e9cdaab4558a2672bb585f76d0ad84e)
Returning to the expression above
|
|
|
(indeterminate form)
|
We apply L'Hôpital's rule once again
![{\displaystyle \ln(k)=\lim _{x\to \infty }{\frac {1}{1}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a9f1c74e27453a264ca7088ab1e0a0e3aa44cf0)
Therefore
![{\displaystyle k=e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c725202c628bba2c779503577e7202d0be08f4f0)
And
![{\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e\neq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e61726bd5138a128ebd536f1f343963d881fe23)
Careful: this does not prove that
because
![{\displaystyle \lim _{x\to \infty }\left(1+{\frac {2}{x}}\right)^{x}=1^{\infty }\neq e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed895eb7cad64c608d24a06cea9a0709d7d0d3ea)
Exercises
Evaluate the following limits using L'Hôpital's rule:
Solutions