L'Hopital's rule

L'Hôpital's Rule is a method in differential calculus for calculating the limit of a quotient of two functions wherein the entire expression approaches an indeterminate form of 0/0 or infinity/infinity. In the event that this is the case, the limit is equal to the limit of the quotient of the first derivatives of the two functions (provided that limit exists). Should this also yield an indeterminate form, the process is repeated until a meaningful result is obtained.

Math form

<math>g'(x)\not\equiv 0</math> C is some number such that <math>f(c)=0</math> <math>g(c)=0</math> <math>\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}</math>

L'Hopital's Rule is not to be confused with the quotient rule, which allows for the calculation of the derivative of a single function that contains a quotient.

Examples

Example 1

A standard application of L'Hopital's rule is in evaluating the limit

<math>\lim_{x \to 0} \frac{\sin x}{x}.</math>

In the preceding notation, this is the situation with <math>f(x) = \sin x</math> and <math>g(x) = x</math>. Both the numerator and the denominator tend to 0 as <math>x</math> tends to 0, i.e., <math>\lim_{x \to 0} \sin x = \lim_{x \to 0} x = 0</math>, and so L'Hôpital's rule implies that

<math>

\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\frac{d}{dx} \sin x}{\frac{d}{dx} x} = \lim_{x \to 0} \frac{\cos x}{1} = \lim_{x \to 0} \cos x = \cos 0 = 1. </math>

Example 2

L'Hopital's rule may also be used in the evaluation of the indeterminate form infinity/infinity. This version of the rule is useful in computing the horizontal asymptotes of rational functions. For example, suppose we seek to compute

<math>\lim_{x \to \infty} \frac{2x^2+3x+2}{x^2-5x+7}.</math>

This is an indeterminate form <math>\infty/\infty</math>. Applying L'Hopital's rule once yields

<math>\lim_{x \to \infty} \frac{2x^2+3x+2}{x^2-5x+7} = \lim_{x \to \infty} \frac{\frac{d}{dx}(2x^2+3x+2)}{\frac{d}{dx}(x^2-5x+7)} = \lim_{x \to \infty} \frac{4x+3}{2x-5}.</math>

This is still an indeterminate form. To evaluate the limit, it is necessary to invoke L'Hopital's rule a second time:

<math>\lim_{x \to \infty} \frac{4x+3}{2x-5} = \lim_{x \to \infty} \frac{\frac{d}{dx}(4x+3)}{\frac{d}{dx}(2x-5)} = \lim_{x \to \infty} \frac{4}{2} = 2.</math>

We conclude that

<math>\lim_{x \to \infty} \frac{2x^2+3x+2}{x^2-5x+7} = 2.</math>

An easy extension of this argument is useful for finding horizontal asymptotes of more general rational functions. Suppose that <math>f</math> and <math>g</math> are two polynomials of equal degree <math>n</math>. Applying L'Hopital's rule <math>n</math> times we may discover that

<math>\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{f_n}{g_n},</math>

where <math>f_n</math> and <math>g_n</math> are the leading coefficients of <math>f</math> and <math>g</math> (i.e., the coefficients on the term <math>x^n</math> in these two polynomials). The example given is a case of this fact with <math>n=2</math> (since both <math>f</math> and <math>g</math> are quadratic), and with <math>f_n = 2</math> and <math>g_n = 1</math>.

Example 3

We can use L'Hôpital's rule to prove the following:

<math>\lim_{x\rightarrow\infty} \frac{x^n}{e^x}=0 \quad\forall n< \infty</math>

This is another example where the limit is in the form of <math>\infty/\infty</math>.
Proof

For Integer n:

<math>\forall n< \infty\quad\lim_{x\rightarrow\infty} \frac{x^n}{e^x}=

                            \lim_{x\rightarrow\infty} \frac{nx^{n-1}}{e^x}=
                            \lim_{x\rightarrow\infty} \frac{n\left(n-1\right)x^{n-2}}{e^x}= \cdots=
                            \lim_{x\rightarrow\infty} \frac{n!}{e^x}= 0</math>

For non-integer n:

<math>\forall n< \infty\quad\lim_{x\rightarrow\infty} \frac{x^n}{e^x}=

                            \lim_{x\rightarrow\infty} \frac{nx^{n-1}}{e^x}=
                            \lim_{x\rightarrow\infty} \frac{n\left(n-1\right)x^{n-2}}{e^x}= \cdots=
                            \lim_{x\rightarrow\infty} \frac{\left[n \left(n-1 \right)\left(n-2 \right)\cdots \left(n-\lfloor n \rfloor \right) \right]x^{n-\lceil n \rceil}}{e^x}</math>

(where <math>\lfloor n \rfloor</math> is the floor function of n and <math>\lceil n \rceil</math> is the ceiling function of n)

Since <math>n-\lceil n \rceil <0</math>, <math>\lim_{x\rightarrow\infty}x^{n-\lceil n \rceil}=0</math> and therefore

<math>\lim_{x\rightarrow\infty} \frac{\left[n \left(n-1 \right)\left(n-2 \right)\cdots \left(n-\lfloor n \rfloor \right) \right]x^{n-\lceil n \rceil}}{e^x}=0</math>

This completes the proof.

Outside Links

• The Complete Idiot's Guide to Calculus