Indeterminate form

In mathematics, a number of the expressions that may be encountered in calculus are considered to be indeterminate forms, and must be treated as symbolic only, until more careful discussion has taken place. The most common one is

${\displaystyle 0/0}$

which has no definite meaning, considering that division by zero is not a meaningful operation in arithmetic. A good page on this is 0/0.

Further examples are

${\displaystyle \infty /\infty }$

${\displaystyle 0\cdot \infty }$

${\displaystyle 1^{\infty }}$

${\displaystyle 0^{0}}$

${\displaystyle \infty ^{0}}$

${\displaystyle \infty -\infty ,}$

all of which are firstly indeterminate forms. If f(x) and g(x) both approach 0 as x approaches some number, or x approaches ∞ or −∞, then

${\displaystyle {f(x) \over g(x)}}$

can approach any real number or ∞ or −∞, or fail to converge to any point on the extended real number line, depending on which functions f and g are; similar remarks are true of the other indeterminate forms displayed above. For example,

${\displaystyle \lim _{x\rightarrow 0}{\sin(x) \over x}=1}$

and

${\displaystyle \lim _{x\rightarrow 49}{x-49 \over {\sqrt {x}}\,-7}=14.}$

Direct substitution of the number that x approaches into either of these functions leads to the indeterminate form 0/0, but both limits actually exist and are 1 and 14 respectively. The indeterminate form does not imply the limit does not exist. In many cases, algebraic elimination, L'Hôpital's rule, or other methods can be used to simplify the expression so the limit can be more easily evaluated.

Logical circularity

In some cases it may constitute circular reasoning to use L'Hopital's rule to evaluate such limits as

${\displaystyle \lim _{h\to 0}{(x+h)^{n}-x^{n} \over h}.}$ 

If one uses the evaluation of the limit above for the purpose of proving that

(d/dx) xⁿ = nxⁿ⁻¹

and one uses L'Hopital's rule and the fact that

(d/dx) xⁿ = nxⁿ⁻¹

in the evaluation of the limit, then one's reasoning is circular and therefore fallacious.