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== Examples ==
== Examples ==
===Example 1===
A standard application of L'Hopital's rule is in evaluating the limit
A standard application of L'Hopital's rule is in evaluating the limit
</math>
</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 [[asymptote|asymptotes]] of rational functions. For example, suppose we seek to compute
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 [[asymptote|asymptotes]] of rational functions. For example, suppose we seek to compute
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>.
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>.<br>
'''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 ==
== Outside Links ==