L'Hôpital's rule
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In calculus, l'Hôpital's rule (alternatively, l'Hospital's rule) uses derivatives to help compute limits with indeterminate forms. Application (or repeated application) of the rule often converts an indeterminate form to a determinate form, allowing easy computation of the limit. The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital, who published the rule in his book Analyse des infiniment petits pour l'intelligence des lignes courbes (1696), the first book to be written on the differential calculus. It is likely, however, that the result was originally due to Johann Bernoulli, upon whose lectures the text was largely based.
Overview
Introduction
In simple cases, L'Hôpital's rule states that:
- [\lim_\frac = \lim_\frac ]
Among other requirements, for this rule to hold, the limit [\lim_\frac] must exist. Other requirements are detailed below, in the formal definition.
Formal definition
When determining the limit of a quotient [f(x)/g(x) \ ] when both f and g approach 0, or f and g approach infinity, L'Hôpital's rule states that [f'/g' \ ] has the same limit (if the limit exists), provided that g′ is nonzero throughout some interval containing the point in question. This differentiation often simplifies the quotient and/or converts it to a determinate form, allowing the limit to be determined more easily.
Symbolically let [\mathbb^*=\mathbb\cup\]. Suppose that [c \in \mathbb^*], that
- [ \lim_ = A, A \in \mathbb^*]
- [\begin \lim_ = \lim_g(x) = 0 \\ \mbox \\ \lim_
>
= +\infty \end] then - [\lim_=A.]
Basic indeterminate forms (all others reduce to these):
[\qquad ]
Other indeterminate forms:
- []
- [\lim_\frac=\lim_(1+\cos(x))]
- [\lim_\frac=1.]
Note also the requirement that the derivative of g not vanish throughout an entire interval containing the point c. Without such a hypothesis, the conclusion is false. Thus one must not use L'Hôpital's rule if the denominator oscillates wildly near the point where one is trying to find the limit. For example if [f(x)=x+\cos(x)\sin(x)] and [g(x)=e^(x+\cos(x)\sin(x))], then
- [\lim_\frac=\frac}\cos(x)(x+\sin(x)\cos(x)+2\cos(x))}
\lim_\frac(x+\sin(x)\cos(x)+2\cos(x))}
whereas
- [\lim_\frac=\lim_\frac}]
Examples
- Here is an example involving the sinc function, which has the form 0/0 :
[\lim_ \mathrm(x)\,] [= \lim_ \frac\,] [= \lim_ \frac\,] [= \lim_ \frac = \frac = 1\,] - However, it is simpler to observe that this limit is just the definition of the derivative of sin(x) at x = 0.
- In fact this particular limit is needed in the most usual proof that the derivative of sin(x) is cos(x), but we cannot use l'Hôpital's rule to do this, as it would produce a circular argument.
- Here is a more elaborate example involving the indeterminate form 0/0. Applying the rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying l'Hôpital's rule three times:
[\lim_ ] [=\lim_] [=\lim_] [=\lim_] [=] [=6\,] - Here is another case involving 0/0:
- :[\lim_=\lim_=\lim_=]
- Here is a case of ∞/∞:
- :[ \lim_ \frac} = \lim_ \frac\,)\ } = \lim_ \frac} = \infty]
- This one involves ∞/∞. Assume n is a positive integer.
- :[\lim_ x^n e^=\lim_=\lim_ \over e^x}=n\lim_ \over e^x}]
- Iterate the above until the exponent is 0. Then one sees that the limit is 0.
- This one also involves ∞/∞:
- :[\lim_ x\ln x=\lim_=\lim_=\lim_ -x = 0]
- This is the impulse response of a raised cosine filter:
[\lim_\, \mathrm(f_0 t)\cdot \frac ] _] [= 1 \cdot 1 = 1] - And:
[\lim_} \mathrm(f_0 t)\cdot \frac ] [= \mathrm\left(\frac\right)\cdot \lim_} \frac] [= \mathrm\left(\frac\right)\cdot \left(\frac\right)] [= \sin\left(\frac\right)\cdot \frac] Proof
The most common proof of l'Hôpital's rule uses Cauchy's mean value theorem.
1) The case when [f(x) \to 0, g(x) \to 0]
First, we expand continuously (or redefine) [f(x)] and [g(x)] by [0] for [x=c]. This doesn't change the limit since the limit doesn't depend on the value in the point (by definition).
According to Cauchy's mean value theorem there is a constant [\xi] in [ c < \xi < c+h ] such that:
- [ = ]
- [ = ]
- [\lim_= \lim_= \lim_ = \lim_]
Let [ x < y < x + h]. Then using Cauchy's mean value theorem:
- [ = ]
- [ = + \left [ 1 - right ] ]
- [\begin\lim_ = B \in \mathbb \\ \\\lim_ = \pm \infty\end]
Other proofs
There are more intuitive proofs of the rule. If
- [\lim_ \frac]
General proof
- [ \lim_ = \lim_]
- [ = ,]
- [ = .]
- [ ~= ,]
- [ \lim_= .]
Other applications
Many other indeterminate forms, such as [1^\infty], [\infty^0], and [\infty-\infty] can be calculated using l'Hôpital's rule.
For example, to handle a case of [\infty-\infty], the difference of two functions is converted to a quotient:
- [ \lim_ x - \sqrt= \lim_ \frac\right) \left(x - \sqrt\right) } } \quad]
- :::[= \lim_ \frac}\quad]
- :::[= \lim_ \frac}\quad]
- :::[= \lim_ \frac}}= \frac = \frac\quad]
Other methods of computing limits
Although l'Hôpital's rule is a powerful way of computing otherwise hard-to-compute limits, it is not always the easiest. Some limits are actually easier to compute using the Taylor series expansion.
For example,
- [ \lim_ x \sin = \lim_ x \left( - + - \cdots \right) \;]
- :::[= \lim_ 1 - + - \cdots\; =\; 1\quad]
- [ \lim_\ \over }]
- [ L = \lim_\ \over }]
- :[= \lim_ } \cdot \cdot ]
- :[= \cos = \cos = 1]
- [Let\ u = ] Therefore, as [ x \rightarrow \infty ], [u \rightarrow\ 0]
- :[\lim_ x \sin \ =\ \lim_ \sin \ =\ 1]
Logical circularity
In some cases it may constitute circular reasoning to use l'Hôpital's rule to evaluate such limits as
- [\lim_.]
- [ x^n=nx^\,]
- [ x^n=nx^\,]
External links
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