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 (e.g. 0/0, 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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)\not\equiv 0} C is some number such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(c)=0} $g(c)=0$ Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} \frac{\sin x}{x}.}

In the preceding notation, this is the situation with $f(x)=\sin x$ and $g(x)=x$. Both the numerator and the denominator tend to 0 as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} tends to 0, i.e., $\lim _{x\to 0}\sin x=\lim _{x\to 0}x=0$, and so L'Hôpital's rule implies that

$\lim _{x\to 0}{\frac {\sin x}{x}}=\lim _{x\to 0}{\frac {{\frac {d}{dx}}\sin x}{{\frac {d}{dx}}x}}=\lim _{x\to 0}{\frac {\cos x}{1}}=\lim _{x\to 0}\cos x=\cos 0=1.$

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

$\lim _{x\to \infty }{\frac {2x^{2}+3x+2}{x^{2}-5x+7}}.$

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

$\lim _{x\to \infty }{\frac {2x^{2}+3x+2}{x^{2}-5x+7}}=\lim _{x\to \infty }{\frac {{\frac {d}{dx}}(2x^{2}+3x+2)}{{\frac {d}{dx}}(x^{2}-5x+7)}}=\lim _{x\to \infty }{\frac {4x+3}{2x-5}}.$

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

$\lim _{x\to \infty }{\frac {4x+3}{2x-5}}=\lim _{x\to \infty }{\frac {{\frac {d}{dx}}(4x+3)}{{\frac {d}{dx}}(2x-5)}}=\lim _{x\to \infty }{\frac {4}{2}}=2.$

We conclude that

$\lim _{x\to \infty }{\frac {2x^{2}+3x+2}{x^{2}-5x+7}}=2.$

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

$\lim _{x\to \infty }{\frac {f(x)}{g(x)}}={\frac {f_{n}}{g_{n}}},$

where $f_{n}$ and $g_{n}$ are the leading coefficients of $f$ and $g$ (i.e., the coefficients on the term $x^{n}$ in these two polynomials). The example given is a case of this fact with $n=2$ (since both $f$ and $g$ are quadratic), and with $f_{n}=2$ and $g_{n}=1$.

Example 3

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow\infty} \frac{x^n}{e^x}=0 \quad\forall n< \infty}

This is another example where the limit is in the form of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \infty/\infty} .
Proof

For Integer n:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}

For non-integer n:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

(where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lfloor n \rfloor} is the floor function of n and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lceil n \rceil} is the ceiling function of n)

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n-\lceil n \rceil <0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow\infty}x^{n-\lceil n \rceil}=0} and therefore

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}

This completes the proof.

Outside Links

• The Complete Idiot's Guide to Calculus