next up previous
Next: Problems Up: No Title Previous: No Title

Cauchy's Mean Value Theorem and L'Hopital's rule

Let f and g be differentiable functions on the interval [a,b], with a < b.
Consider the following function:

h(x) = (f(x) - f(a)(g(b) - g(a)) - (g(x) - g(a))(f(b) - f(a).

Then if we evaluate h at x = a and at x = b, we get:

h(a) = h(b) = 0.

So by Rolle's theorem there exists a point $\xi$ in the interval (a, b), such that $h'(\xi) = 0$.
Then we have $0 = h'(\xi) = f'(\xi)(g(b) - g(a)) - g'(\xi)(f(b) - f(a))$.
In particular, if g' is non-zero in the open interval (a,b) and if $g(b) \ne g(a)$, we have, for some $\xi$ in the interval (a,b):

\begin{displaymath}\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\end{displaymath}

This is called Cauchy's Mean Value Theorem.
In the special case that g(x) = x, so g'(x) = 1, this reduces to the ordinary mean value theorem.
Now consider the case that both f(a) and g(a) vanish and replace b by a variable x.
Then we have, provided f(a) = g(a) = 0 and $g'(x) \ne 0$ in an interval around a, except possibly at x = a:

\begin{displaymath}\frac{f(x)}{g(x)} = \frac{f'(\xi(x))}{g'(\xi(x))},\end{displaymath}

where $x \ne a$, but x is near a and then $\xi(x)$ (which depends on x) is a point between x and a.
Now suppose $\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L$ exists.
Then $\lim_{x\rightarrow a} \frac{f'(\xi(x))}{g'(\xi(x))} = L$ also, so we get L'Hopital's rule:

\begin{displaymath}\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}.\end{displaymath}

It must be emphasized that this rule only has been shown to hold when both f(a) and g(a) vanish.
If now f and g are twice differentiable and f'(a) and g'(a) both vanish and if g''(x) is non-zero near x = a, except possibly at x = a and then if $\lim_{x\rightarrow a}
\frac{f''(x)}{g''(x)}$ exists, we may apply L'Hopital's rule twice to give:

\begin{displaymath}\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow ...
...rac{f'(x)}{g'(x)} = \lim_{x\rightarrow a}\frac{f''(x)}{g''(x)}.\end{displaymath}

Finally if g and its first n-1-derivatives vanish at a, and if g and its first n derivatives do not vanish in an interval around a, except possibly at a, then we have the iterated L'Hopital's rule:

\begin{displaymath}\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f^{(n)}(x)}{g^{(n)}(x)}.\end{displaymath}

Examples:

\begin{displaymath}\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = \lim_{x\rightarrow 0} \frac{\cos(x)}{1} = 1, \end{displaymath}


\begin{displaymath}\lim_{x\rightarrow 0}\frac{\sin(x) - x}{x^3} = \lim_{x\righta...
...2} = \lim_{x\rightarrow 0}
\frac{-\sin(x)}{6x} = -\frac{1}{6}. \end{displaymath}

L'Hopital's rule is also valid for other cases, although we do not prove these: Examples:

\begin{displaymath}\lim_{x\rightarrow \infty} x^3e^{-x} = \lim_{x\rightarrow \in...
... \frac{6x}{e^x} = \lim_{x\rightarrow \infty} \frac{6}{e^x} = 0.\end{displaymath}


\begin{displaymath}\lim_{x\rightarrow \infty} x(e^{\frac{1}{x}} - 1) = \lim_{x\r...
...= \lim_{x\rightarrow \infty}
(- e^{\frac{1}{x}}) = - e^0 = - 1.\end{displaymath}


\begin{displaymath}\lim_{x\rightarrow \infty} \frac{(\tan^{-1}(x) - \frac{\pi}{2...
...\lim_{x\rightarrow \infty}
- \frac{1}{1 + \frac{1}{x^2}} = - 1.\end{displaymath}


\begin{displaymath}\lim_{x\rightarrow \infty} x(\ln(x + 5) - \ln(x)) = \lim_{x\rightarrow \infty} \frac{\ln(\frac{x + 5}{x})}{\frac{1}{x}}\end{displaymath}


\begin{displaymath}= 
 \lim_{x\rightarrow \infty} \frac{\ln(1 + \frac{5}{x})}{\f...
...}} = \lim_{x\rightarrow \infty} 5\frac{1}{1 +
\frac{5}{x}} = 5.\end{displaymath}

Sometimes the calculation goes easier if we replace x by $\frac{1}{x}$.
Then a limit to $\infty$ is replaced by a limit to 0+ and a limit to $-\infty$ is replaced by a limit to 0- and vice-versa.
Using this trick the last example reads:

\begin{displaymath}\lim_{x\rightarrow \infty} x(\ln(x + 5) - \ln(x)) = \lim_{x\rightarrow \infty} \frac{\ln(\frac{x + 5}{x})}{\frac{1}{x}}\end{displaymath}


\begin{displaymath}= 
 \lim_{x\rightarrow \infty} \frac{\ln(1 + \frac{5}{x})}{\f...
...{x} =
\lim_{x\rightarrow 0^+} \frac{(\frac{5}{1 + 5x})}{1} = 5.\end{displaymath}

Next we can handle some exponent limits by first taking logarithms and then using L'Hopital.
Consider $\lim_{x\rightarrow \infty}(1 + \frac{1}{x})^x$.
Let $y = (1 + \frac{1}{x})^x$, then $\ln(y) = x\ln(1 + \frac{1}{x})$ and we have:

\begin{displaymath}\lim_{x\rightarrow \infty} \ln(y) = \lim_{x\rightarrow \infty...
... \frac{1}{x}) = \lim_{x\rightarrow 0^+}(\frac{1}{x}\ln(1 +x)) 
\end{displaymath}


\begin{displaymath}= \lim_{x\rightarrow 0^+}(\frac{\ln(1 +x)}{x}) = \lim_{x\rightarrow 0^+}(\frac{(\frac{1}{1 + x})}{1}) = 1.\end{displaymath}

So $\lim_{x\rightarrow \infty}(1 + \frac{1}{x})^x = \lim_{x\rightarrow \infty}e^{\ln(y)} = e^{\lim_{x\rightarrow
\infty}(\ln(y))} = e^1 = e$.

Finally, although again we do no prove this, L'Hopital can be used to show that limits do not exist. For example:

\begin{displaymath}\lim_{x \rightarrow 0}\frac{\sin(x) - x}{x^4} = \lim_{x \rightarrow 0}\frac{\cos(x) - 1}{4x^3}\end{displaymath}


\begin{displaymath}= \lim_{x \rightarrow
0}\frac{-\sin(x)}{12x^2} = \lim_{x \rightarrow
0}\frac{-\cos(x)}{24x},\end{displaymath}

which does not exist, since the numerator goes to -1 and the denominator goes to zero.

 
next up previous
Next: Problems Up: No Title Previous: No Title
George A. J. Sparling
2001-01-11