Suppose a function f(x) is of the form f(x) = $\frac{g(x)}{h(x)}$. When both g(a) and h(a) are equal to zero, then $\lim_{x\rightarrow a}f(x)$ = $\lim_{x\rightarrow a}\frac{g(x)}{h(x)}$ assumes the form $\frac{0}{0}$ when g(a) and h(a) are directly substituted in the limit. The limit may nevertheless exist, and can be computed using some algebraic manipulations. Hence the expressions $\frac{0}{0}$ or for that matter $\frac{\infty }{\infty }$ are used as notations to represent an indeterminate form.
Example: Let f(x) = $\frac{x^{2}-4}{x-2}$. The limit of f(x) as x $\rightarrow$ 2 becomes on substitution $\frac{0}{0}$ representing the indeterminate form of the limit.But the limit can be found by factoring the numerator and canceling the zero yielding factor (x - 2) with the denominator as follows.
$\lim_{x\rightarrow 2}\frac{x^{2}-4}{x-2}$ = $\lim_{x\rightarrow a}\frac{(x+2)(x-2)}{(x-2)}$ = $\lim_{x\rightarrow 2}(x+2)$ = 4

But not all functions can be manipulated algebraically to determine the limit if it exists. For example consider the function g(x) = $\frac{ln(x-1)}{x-2}$. We get $\frac{0}{0}$ indeterminate for the limit $\lim_{x\rightarrow 2}g(x)$ on substitution. The limit cannot be evaluated using known algebraic techniques of cancellation, rationalization etc.

The French mathematician Guillaume de L'Hospital first published a rule for evaluating these indeterminate forms using differentiation. The rule of applying the limit on successive derivatives of the numerator and denominator is popularly known as L'Hospital's Rule.

## List of Indeterminate Forms

Let us look at the statement of L'Hospital's Rule and how it is applied to find a limit expressed in indeterminate form.

Suppose f(x) and g(x) are differentiable and g'(x) ≠ 0, near x = a.

If $\lim_{x\rightarrow a}f(x)$ = 0 and $\lim_{x\rightarrow a}g(x)$ = 0 or equivalently
$\lim_{x\rightarrow a}f(x)$ = ±∞ and $\lim_{x\rightarrow a}g(x)$ = $\pm$ $\infty$
(This means indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$), then
$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ = $\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$
Again if $\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$ exists are +$\infty$ or -$\infty$.
The types of indeterminate forms other than these are listed as follows:
1. Product of 0 and infinity -> 0 . $\infty$ form.
2. Zero raised to zero -> 00 form
3. Infinity raised to zero -> $\infty^o$ form
4. One raised to infinity -> 1 * $\infty$ form
5. Difference of infinity -> $\infty$ - $\infty$ form.

Limits in the above list of indeterminate forms can be mathematically manipulated and be written in either $\frac{0}{0}$ form or in $\frac{\infty }{\infty }$ form and then L'Hospital's Rule can be applied to evaluate the limit.

## Indeterminate Forms Examples

### Solved Examples

Question 1: Find $\lim_{x\rightarrow 2}\frac{ln(x-1)}{(x-2)}$
Solution:

The given limit is of the indeterminate form of the type $\frac{0}{0}$   because on substitution we get
$\lim_{x\rightarrow 2}ln(x-1)$ = ln (2 -1) = ln(1) =0    and
$\lim_{x\rightarrow 2}(x-2)$ = 0.
Hence L'Hospital's Rule can be applied and we have

$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ = $\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$

$\lim_{x\rightarrow 2}\frac{ln(x-1)}{(x-2)}$ = $\lim_{x\rightarrow 2}\frac{\frac{1}{x-1}}{1}$ = $\frac{1}{1}$ = 1

Question 2: Evaluate $\lim_{x\rightarrow \infty }\frac{e^{x}}{x}$
Solution:

The given limit is of the form $\frac{\infty }{\infty }$
Hence L'Hospital's Rule can be applied directly as follows
$\lim_{x\rightarrow \infty }\frac{e^{x}}{x}$ = $\lim_{x\rightarrow \infty }\frac{e^{x}}{1}$ = .

Question 3: Evaluate $\lim_{x\rightarrow 0}\frac{x^{2}}{e^{x}-x}$
Solution:

The given limit is of $\frac{0}{0}$ form
Applying L'Hospital's Rule we get
$\lim_{x\rightarrow 0}\frac{x^{2}}{e^{x}-x}$ = $\lim_{x\rightarrow 0}\frac{2x}{e^{x}-1}$
The limit resulted from the application of the rule is again of $\frac{0}{0}$ form.
Hence as the conditions for L'Hospital's rule apply, we continue the process and get
$\lim_{x\rightarrow 0}\frac{2x}{e^{x}-1}$ = $\lim_{x\rightarrow 0}\frac{2}{e^{x}}$ = $\frac{2}{\infty }$ = 0.

Question 4: Find the limit of $\lim_{x\rightarrow 0}x.ln(x)$.
Solution:

On substitute we get the indeterminate form of type 0 . ∞.  The expression needs to be rewritten, so that we can apply L'Hospital's Rule for finding the limit.

$\lim_{x\rightarrow 0}x.ln(x)$ = $\lim_{x\rightarrow 0}\frac{ln(x)}{\frac{1}{x}}$

The resulting limit is of indetermiante form $\frac{\infty }{\infty }$

Hence applying L'Hospital's rule we get,

$\lim_{x\rightarrow 0}x.ln(x)$ = $\lim_{x\rightarrow 0}\frac{ln(x)}{\frac{1}{x}}$

= $\lim_{x\rightarrow 0}\frac{\frac{1}{x}}{\frac{-1}{x^{2}}}$

= $\lim_{x\rightarrow 0}-x$ = 0