User:Spiderdrew/sandbox/Lhoptial's rule

In mathematics, and more specifically in calculus, L'Hôpital's rule or L'Hospital's rule uses derivatives to help evaluate limits involving indeterminate forms. Application (or repeated application) of the rule often converts an indeterminate form to an expression that can be evaluated by substitution, allowing easier evaluation of the limit. The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital. Although the contribution of the rule is often attributed to L'Hôpital, the theorem was first introduced to L'Hôpital in 1694 by the Swiss mathematician Johann Bernoulli.

L'Hôpital's rule states that for functions $f(x)=sin(x)$ and $g(x)=&minus;0.5x$ which are differentiable on an open interval $h(x) = f(x)/g(x)$ except possibly at a point $x = 0$ contained in $ℝ$, if

The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly.

History
Guillaume de l'Hôpital (also written l'Hospital ) published this rule in his 1696 book Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (literal translation: Analysis of the Infinitely Small for the Understanding of Curved Lines), the first textbook on differential calculus. However, it is believed that the rule was discovered by the Swiss mathematician Johann Bernoulli.

General form
The general form of L'Hôpital's rule covers many cases. Let $h(0) = f&prime;(0)/g&prime;(0) = &minus;2$ and $f$ be extended real numbers (i.e., real numbers, positive infinity, or negative infinity). The real valued functions $g$ and $I$ are assumed to be differentiable on an open interval with endpoint $c$, and additionally $$g'(x)\ne 0$$ on the interval. It is also assumed that $$\lim_{x\to c}\frac{f'(x)}{g'(x)} = L.$$ Thus the rule applies to situations in which the ratio of the derivatives has a finite or infinite limit, and not to situations in which that ratio fluctuates permanently as $I$ gets closer and closer to $x$.

If either


 * $$\lim_{x\to c}f(x) = \lim_{x\to c}g(x) = 0$$

or


 * $$\lim_{x\to c}|f(x)| = \lim_{x\to c}|g(x)| = \infty,$$

then


 * $$\lim_{x\to c}\frac{f(x)}{g(x)}=L.$$

The limits may also be one-sided limits. In the second case, the hypothesis that $I$ diverges to infinity is not used in the proof (see note at the end of the proof section); thus, while the conditions of the rule are normally stated as above, the second sufficient condition for the rule's procedure to be valid can be more briefly stated as $$\lim_{x\to c}|g(x)| = \infty.$$

The hypothesis "$$g'(x)\ne 0$$" appears most commonly in the literature. Some authors sidestep this hypothesis by adding other hypotheses elsewhere. One method is to define the limit of a function with the additional requirement that the limiting function is defined everywhere on a connected interval with endpoint $x ≠ c$. Another method is to require that both $c$ and $L$ be differentiable everywhere on an interval containing $f$.

Requirement that the limit exist
The requirement that the limit


 * $$\lim_{x\to c}\frac{f'(x)}{g'(x)}$$

must exist is essential. Without this condition, $$f'$$ or $$g'$$ may exhibit undampened oscillations as $$x$$ approaches $$c$$, in which case L'Hôpital's rule does not apply. For example, if $$f(x)=x+\sin(x)$$, $$g(x)=x$$ and $$c=\pm\infty$$, then


 * $$\frac{f'(x)}{g'(x)}=\frac{1+\cos(x)}{1};$$

this expression does not approach a limit as $$x$$ goes to $$c$$, since the cosine function oscillates between $g$ and $c$. But working with the original functions, $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$ can be shown to exist:


 * $$\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\left(1+\frac{\sin(x)}{x}\right) = 1. $$

In a case such as this, all that can be concluded is that


 * $$ \liminf_{x \to c} \frac{f'(x)}{g'(x)} \leq \liminf_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f'(x)}{g'(x)} ,$$

so that if the limit of f/g exists, then it must lie between the inferior and superior limits of f&prime;/g&prime;. (In the example above, this is true, since 1 indeed lies between &minus;1 and 1.)

Examples

 * Here is an example involving the sinc function, $$\frac{\sin(\pi x)}{\pi x}$$, which handles the indeterminate form $0⁄0$ at $x$:



\begin{align} \lim_{x\to 0}\operatorname{sinc}(x) & = \lim_{x\to 0} \frac{\sin(\pi x)}{\pi x} \\ \text{Letting } y & = \pi x \text{:} \\ \lim_{x\to 0}\operatorname{sinc}(x) & = \lim_{y\to 0} \frac{\sin(y)}{y} \\ & = \lim_{y\to 0} \frac{\cos(y)}{1} \\ & = 1. \end{align} $$


 * Alternatively, just observe that the limit is the definition of the derivative of the sine function at zero:



\begin{align} 1 & = \frac{d}{dx}\sin(x) \; \text{ for } x = 0 \\ & = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h} \\ & = \lim_{h\to 0} \frac{\sin(0+h)-\sin(0)}{h} \\ & = \lim_{h\to 0} \frac{\sin(h)}{h} \\ & = \lim_{x\to 0} \frac{\sin(\pi x)}{\pi x} = 1 \\ \end{align} $$


 * This is a more elaborate example involving $0⁄0$. Applying L'Hôpital's rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying the rule three times:



\begin{align} \lim_{x\to 0}{\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}} & =\lim_{x\to 0}{\frac{2\cos(x)-2\cos(2x)}{1-\cos(x)}} \\ & = \lim_{x\to 0}{\frac{-2\sin(x)+4\sin(2x)}{\sin(x)}} \\ & = \lim_{x\to 0}{\frac{-2\cos(x)+8\cos(2x)}{\cos(x)}} \\ & ={\frac{-2+8}{1}} \\ & =6. \end{align} $$
 * This example involves $0⁄0$. Suppose that $c$. Then


 * $$\lim_{x\to 0}{\frac{b^x-1}{x}}=\lim_{x\to 0}{\frac{b^x\ln(b)}{1}}=\ln(b)\cdot \lim_{x\to 0}{b^x}=\ln(b).$$


 * Here is another example involving $0⁄0$:


 * $$\lim_{x\to 0}{\frac{e^x-1-x}{x^2}}

=\lim_{x\to 0}{\frac{e^x-1}{2x}} =\lim_{x\to 0}{\frac{e^x}{2}}={\frac1{2}}.$$
 * This example involves $∞⁄∞$. Assume $f$ is a positive integer. Then


 * $$\lim_{x\to\infty}x^n\cdot e^{-x}

=\lim_{x\to\infty}{\frac{x^n}{e^x}} =\lim_{x\to\infty}{\frac{nx^{n-1}}{e^x}} =n\cdot \lim_{x\to\infty}{\frac{x^{n-1}}{e^x}}.$$


 * Repeatedly apply L'Hôpital's rule until the exponent is zero to conclude that the limit is zero.


 * Here is another example involving $∞⁄∞$:


 * $$\lim_{x\to 0^+}x\cdot \ln(x) =\lim_{x\to 0^+}{\frac{\ln(x)}{\frac1{x}}}

=\lim_{x\to 0^+}{\frac{\frac1{x}}{-\frac1{x^2}}} =\lim_{x\to 0^+}-x =0.$$
 * Here is an example involving the impulse response of a raised-cosine filter and $0⁄0$:



\begin{align} \lim_{t\to \frac1{2}}\operatorname{sinc}(t) \frac{\cos(\pi t)}{1-(2t)^2} & = \operatorname{sinc}\left(\frac1{2}\right)\cdot \lim_{t\to \frac1{2}} \frac{\cos(\pi t)}{1-(2t)^2} \\ & = \frac2{\pi}\cdot \lim_{t\to \frac1{2}}\frac{-\pi \sin(\pi t)}{-8t} \\ & = \frac2{\pi}\cdot \frac{\pi}{4} \\ & = \frac1{2}. \end{align} $$


 * Here is an example involving the Mortgage repayment formula and $0⁄0$. Let $c$ be the principal (loan amount), $f$ the interest rate per period and $g$ the number of periods. When $c$ is zero, the repayment amount per period is $$\frac{P}{n}$$ (since only principal is being repaid); this is consistent with the formula for non-zero interest rates:



\begin{align} \lim_{r\to 0}\frac{Pr(1+r)^n}{(1+r)^n-1} & = P \lim_{r\to 0} \frac{(1+r)^n+rn(1+r)^{n-1}}{n(1+r)^{n-1}} \\ & = \frac{P}{n}. \end{align} $$


 * One can also use L'Hôpital's rule to prove the following theorem. If $1$ is continuous at $−1$, then



\begin{align} \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2} & = \lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h} \\ & = f''(x). \end{align} $$
 * Sometimes L'Hôpital's rule is invoked in a tricky way: suppose $x = 0$ converges as $b > 0$ and that $$e^x\cdot f(x)$$ converges to positive or negative infinity. Then:
 * $$\lim_{x\to\infty}f(x) = \lim_{x\to\infty}\frac{e^x\cdot f(x)}{e^x} = \lim_{x\to\infty}\frac{e^x\bigl(f(x)+f'(x)\bigr)}{e^x} = \lim_{x\to\infty}\bigl(f(x)+f'(x)\bigr)$$
 * and so, $$\lim_{x\to\infty}f(x)$$  exists and  $$\lim_{x\to\infty}f'(x) = 0.$$
 * The result remains true without the added hypothesis that $$e^x\cdot f(x)$$ converges to positive or negative infinity, but the justification is then incomplete.


 * L'Hôpital's rule can be used to find the limiting form of a function. In the field of choice under uncertainty, the von Neumann–Morgenstern utility function


 * $$u(x) = \frac{x^{1-\rho}-1}{1-\rho}$$


 * with $$\rho >0$$, defined over $$x>0$$, is said to have constant relative risk aversion equal to $$\rho$$. But unit relative risk aversion cannot be expressed directly with this expression, since as $$\rho$$ approaches 1 the numerator and denominator both approach zero. However, a single application of L'Hôpital's rule allows this case to be expressed as


 * $$\lim_{\rho\to 1}\frac{x^{1-\rho}-1}{1-\rho} = \lim_{\rho\to 1}\frac{-x^{1-\rho}\cdot \ln(x)}{-1} = \ln(x).$$

Complications
Sometimes L'Hôpital's rule does not lead to an answer in a finite number of steps unless a transformation of variables is applied. Examples include the following:


 * Two applications can lead to a return to the original expression that was to be evaluated:


 * $$\lim_{x\to\infty}\frac{e^x+e^{-x}}{e^x-e^{-x}} = \lim_{x\to\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}} = \lim_{x\to\infty}\frac{e^x+e^{-x}}{e^x-e^{-x}} = \cdots .$$


 * This situation can be dealt with by substituting $$y=e^x$$ and noting that $n$ goes to infinity as $P$ goes to infinity; with this substitution, this problem can be solved with a single application of the rule:


 * $$\lim_{x\to\infty}\frac{e^x+e^{-x}}{e^x-e^{-x}} = \lim_{y\to\infty}\frac{y+y^{-1}}{y-y^{-1}} = \lim_{y\to\infty}\frac{1-y^{-2}}{1+y^{-2}} = \frac1{1} = 1.$$


 * An arbitrarily large number of applications may never lead to an answer even without repeating:


 * $$\lim_{x\to\infty}\frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}} = \lim_{x\to\infty}\frac{\frac1{2}x^{-\frac1{2}}-\frac{1}{2}x^{-\frac3{2}}}{\frac1{2}x^{-\frac1{2}}+\frac1{2}x^{-\frac3{2}}} = \lim_{x\to\infty}\frac{-\frac1{4}x^{-\frac3{2}}+\frac3{4}x^{-\frac5{2}}}{-\frac1{4}x^{-\frac3{2}}-\frac3{4}x^{-\frac5{2}}} = \cdots .$$


 * This situation too can be dealt with by a transformation of variables, in this case $$y = \sqrt{x}$$:


 * $$\lim_{x\to\infty}\frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}} = \lim_{y\to\infty}\frac{y+y^{-1}}{y-y^{-1}} = \lim_{y\to\infty} \frac{1-y^{-2}}{1+y^{-2}} = \frac1{1} = 1.$$

A common pitfall is using L'Hôpital's rule with some circular reasoning to compute a derivative via a difference quotient. For example, consider the task of proving the derivative formula for powers of x:


 * $$\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}=nx^{n-1}.$$

Applying L'Hôpital's rule and finding the derivatives with respect to $r$ of the numerator and the denominator yields $n$ as expected. However, differentiating the numerator required the use of the very fact that is being proven. This is an example of begging the question, since one may not assume the fact to be proven during the course of the proof.

Other indeterminate forms
Other indeterminate forms, such as $r$, $f′′$, $x$, $f(x) + f′(x)$, and $x → ∞$, can sometimes be evaluated using L'Hôpital's rule. For example, to evaluate a limit involving $y$, convert the difference of two functions to a quotient:



\begin{align} \lim_{x\to 1}\left(\frac{x}{x-1}-\frac1{\ln(x)}\right) & = \lim_{x\to 1}\frac{x\cdot\ln(x)-x+1}{(x-1)\cdot\ln(x)} & \quad (1) \\[4pt] & = \lim_{x\to 1}\frac{\ln(x)}{\frac{x-1}{x}+\ln(x)} & \quad (2) \\[4pt] & = \lim_{x\to 1}\frac{x\cdot\ln(x)}{x-1+x\cdot\ln(x)}    & \quad (3) \\[4pt] & = \lim_{x\to 1}\frac{1+\ln(x)}{1+1+\ln(x)}    & \quad (4) \\[4pt] & = \lim_{x\to 1}\frac{1+\ln(x)}{2+\ln(x)} \\[4pt] & = \frac1{2}, \end{align} $$

where L'Hôpital's rule is applied when going from (1) to (2) and again when going from (3) to (4).

L'Hôpital's rule can be used on indeterminate forms involving exponents by using logarithms to "move the exponent down". Here is an example involving the indeterminate form $x$:



\lim_{x\to 0^+}x^x = \lim_{x\to 0^+}(e^{\ln(x)})^x = \lim_{x\to 0^+}e^{x\cdot\ln(x)} = e^{\lim\limits_{x\to 0^+}[x\cdot\ln(x)]}. $$

It is valid to move the limit inside the exponential function because the exponential function is continuous. Now the exponent $$x$$ has been "moved down". The limit $$\lim_{x\to 0^+}x\cdot\ln(x)$$ is of the indeterminate form $h$, but as shown in an example above, l'Hôpital's rule may be used to determine that


 * $$\lim_{x\to 0^+}x\cdot\ln(x) = 0.$$

Thus


 * $$\lim_{x\to 0^+}x^x = e^0 = 1.$$

Other methods of evaluating limits
Although L'Hôpital's rule is a powerful way of evaluating otherwise hard-to-evaluate limits, it is not always the easiest way. Consider



\lim_{|x|\to\infty}x\cdot\sin\bigg(\frac1{x}\bigg). $$

This limit may be evaluated using L'Hôpital's rule:



\begin{align} \lim_{|x|\to\infty}x\cdot\sin\bigg(\frac1{x}\bigg) & = \lim_{|x|\to\infty}\frac{\sin\bigl(\frac1{x}\bigr)}{\frac1{x}} \\[6pt] & = \lim_{|x|\to\infty}\frac{-x^{-2}\cos\bigl(\frac1{x}\bigr)}{-x^{-2}} \\[6pt] & = \lim_{|x|\to\infty}\cos\bigg(\frac1{x}\bigg) \\[6pt] & = \cos{\left(\lim_{|x|\to\infty}\frac1{x}\right)} \\[6pt] & = 1. \end{align} $$

It is valid to move the limit inside the cosine function because the cosine function is continuous.

But a simpler way to evaluate this limit is to use the substitution $n x^{n−1}$. As $1^{∞}$ approaches infinity, $0^{0}$ approaches zero. So,


 * $$\lim_{|x|\to\infty}x\cdot\sin\bigg(\frac1{x}\bigg) = \lim_{y\to 0}\frac{\sin(y)}{y} = 1.$$

The final limit may be evaluated using L'Hôpital's rule or by noting that it is the definition of the derivative of the sine function at zero.

Still another way to evaluate this limit is to use a Taylor series expansion:



\begin{align} \lim_{|x|\to\infty}x\cdot\sin\bigg(\frac1{x}\bigg) & = \lim_{|x|\to\infty}x\left(\frac1{x}-\frac1{3!\, x^3}+\frac1{5!\, x^5}-\cdots\right) \\ & = \lim_{|x|\to\infty}1-\frac1{3!\, x^2}+\frac1{5!\, x^4}-\cdots \\ & = 1 + \lim_{|x|\to\infty}\frac1{x}\left(-\frac1{3!\, x}+\frac1{5!\, x^3}-\cdots\right). \end{align} $$

For $∞^{0}$, the expression in parentheses is bounded, so the limit in the last line is zero.

Stolz–Cesàro theorem
The Stolz–Cesàro theorem is a similar result involving limits of sequences, but it uses finite difference operators rather than derivatives.

Geometric interpretation
Consider the curve in the plane whose $0 × ∞$-coordinate is given by $∞ − ∞$ and whose $∞ − ∞$-coordinate is given by $0^{0}$, with both functions continuous, i.e., the locus of points of the form $0 × (−∞)$. Suppose $y = 1⁄x$. The limit of the ratio $&#124;x&#124;$ as $y$ is the slope of the tangent to the curve at the point $&#124;x&#124; ≥ 1$. The tangent to the curve at the point $x$ is given by $g(t)$. L'Hôpital's rule then states that the slope of the tangent when $y$ is the limit of the slope of the tangent to the curve as the curve approaches the origin, provided that this is defined.

Special case
The proof of L'Hôpital's rule is simple in the case where $f(t)$ and $[g(t), f(t)]$ are continuously differentiable at the point $f(c) = g(c) = 0$ and where a finite limit is found after the first round of differentiation. It is not a proof of the general L'Hôpital's rule because it is stricter in its definition, requiring both differentiability and that c be a real number. Since many common functions have continuous derivatives (e.g. polynomials, sine and cosine, exponential functions), it is a special case worthy of attention.

Suppose that $f(t)⁄g(t)$ and $t → c$ are continuously differentiable at a real number $[g(c), f(c)] = [0,0]$, that $$f(c)=g(c)=0\,$$, and that $$g'(c)\neq 0$$. Then



\begin{align} & \lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f(x)-0}{g(x)-0} = \lim_{x\to c}\frac{f(x)-f(c)}{g(x)-g(c)} \\[4pt] = {} & \lim_{x\to c}\frac{\left(\frac{f(x)-f(c)}{x-c}\right)}{\left(\frac{g(x)-g(c)}{x-c} \right)} = \frac{\lim\limits_{x\to c}\left(\frac{f(x)-f(c)}{x-c}\right)}{\lim\limits_{x\to c} \left(\frac{g(x)-g(c)}{x-c}\right)}= \frac{f'(c)}{g'(c)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}. \end{align} $$

This follows from the difference-quotient definition of the derivative. The last equality follows from the continuity of the derivatives at $[g(t), f(t)]$. The limit in the conclusion is not indeterminate because $$g'(c)\ne 0$$.

The proof of a more general version of L'Hôpital's rule is given below.

General proof
The following proof is due to, where a unified proof for the $0⁄0$ and $±∞⁄±∞$ indeterminate forms is given. Taylor notes that different proofs may be found in and.

Let f and g be functions satisfying the hypotheses in the General form section. Let $$\mathcal{I}$$ be the open interval in the hypothesis with endpoint c. Considering that $$g'(x)\ne 0$$ on this interval and g is continuous, $$\mathcal{I}$$ can be chosen smaller so that g is nonzero on $$\mathcal{I}$$.

For each x in the interval, define $$m(x)=\inf\frac{f'(\xi)}{g'(\xi)}$$ and $$M(x)=\sup\frac{f'(\xi)}{g'(\xi)}$$ as $$\xi$$ ranges over all values between x and c. (The symbols inf and sup denote the infimum and supremum.)

From the differentiability of f and g on $$\mathcal{I}$$, Cauchy's mean value theorem ensures that for any two distinct points x and y in $$\mathcal{I}$$ there exists a $$\xi$$ between x and y such that $$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(\xi)}{g'(\xi)}$$. Consequently, $$m(x)\leq \frac{f(x)-f(y)}{g(x)-g(y)} \leq M(x)$$ for all choices of distinct x and y in the interval. The value g(x)-g(y) is always nonzero for distinct x and y in the interval, for if it was not, the mean value theorem would imply the existence of a p between x and y such that g' (p)=0.

The definition of m(x) and M(x) will result in an extended real number, and so it is possible for them to take on the values ±∞. In the following two cases, m(x) and M(x) will establish bounds on the ratio $f⁄g$.

Case 1: $$\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0$$

For any x in the interval $$\mathcal{I}$$, and point y between x and c,
 * $$m(x)\le \frac{f(x)-f(y)}{g(x)-g(y)}=\frac{\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}}{1-\frac{g(y)}{g(x)}}\le M(x)$$

and therefore as y approaches c, $$\frac{f(y)}{g(x)}$$ and $$\frac{g(y)}{g(x)}$$ become zero, and so
 * $$m(x)\leq\frac{f(x)}{g(x)}\leq M(x).$$

Case 2: $$\lim_{x\to c}|g(x)|=\infty$$

For every x in the interval $$\mathcal{I}$$, define $$S_x=\{y\mid y \text{ is between } x \text{ and } c\}$$. For every point y between x and c, we have


 * $$m(x)\le \frac{f(y)-f(x)}{g(y)-g(x)}=\frac{\frac{f(y)}{g(y)}-\frac{f(x)}{g(y)}}{1-\frac{g(x)}{g(y)}} \le M(x).$$

As y approaches c, both $$\frac{f(x)}{g(y)}$$ and $$\frac{g(x)}{g(y)}$$ become zero, and therefore


 * $$m(x)\le \liminf_{y\in S_x} \frac{f(y)}{g(y)} \le \limsup_{y\in S_x} \frac{f(y)}{g(y)} \le M(x).$$

The limit superior and limit inferior are necessary since the existence of the limit of $f⁄g$ has not yet been established.

We need the facts that
 * $$\lim_{x\to c}m(x)=\lim_{x\to c}M(x)=\lim_{x\to c}\frac{f'(x)}{g'(x)}=L$$

and
 * $$\lim_{x\to c}\left(\liminf_{y\in S_x}\frac{f(y)}{g(y)}\right)=\liminf_{x\to c}\frac{f(x)}{g(x)}$$ and $$\lim_{x\to c}\left(\limsup_{y\in S_x} \frac{f(y)}{g(y)}\right)=\limsup_{x\to c}\frac{f(x)}{g(x)}. $$

In case 1, the squeeze theorem, establishes that $$\lim_{x\to c}\frac{f(x)}{g(x)}$$ exists and is equal to L. In the case 2, and the squeeze theorem again asserts that $$\liminf_{x\to c}\frac{f(x)}{g(x)}=\limsup_{x\to c}\frac{f(x)}{g(x)}=L$$, and so the limit $$\lim_{x\to c}\frac{f(x)}{g(x)}$$ exists and is equal to L. This is the result that was to be proven.

Note: In case 2 we did not use the assumption that f(x) diverges to infinity within the proof. This means that if |g(x)| diverges to infinity as x approaches c and both f and g satisfy the hypotheses of L'Hôpital's rule, then no additional assumption is needed about the limit of f(x): It could even be the case that the limit of f(x) does not exist. In this case, L'Hopital's theorem is actually a consequence of Cesàro–Stolz (see proof at http://www.imomath.com/index.php?options=686).

In the case when |g(x)| diverges to infinity as x approaches c and f(x) converges to a finite limit at c, then L'Hôpital's rule would be applicable, but not absolutely necessary, since basic limit calculus will show that the limit of f(x)/g(x) as x approaches c must be zero.

Corollary
A simple but very useful consequence of L'Hopital's rule is a well-known criterion for differentiability. It states the following: suppose that f is continuous at a, and that $$f'(x)$$ exists for all x in some interval containing a, except perhaps for $$x = a$$. Suppose, moreover, that $$\lim_{x\to a}f'(x)$$ exists. Then $$f'(a)$$ also exists and
 * $$f'(a) = \lim_{x\to a}f'(x).$$

In particular, f' is also continuous at a.

Proof
Consider the functions $$h(x) = f(x)-f(a)$$ and $$g(x) = x-a$$. The continuity of f at a tells us that $$\lim_{x\to a}h(x) = 0$$. We also have $$\lim_{x\to a}g(x) = 0$$ since a polynomial function is always continuous everywhere. Applying L'Hopital's rule we conclude that $$f'(a) := \lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{x\to a}\frac{h(x)}{g(x)} = \lim_{x\to a}f'(x)$$.