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.

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.

Solved Examples

Question 1: Find $\lim_{x\rightarrow 2}\frac{ln(x-1)}{(x-2)}$
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 $\lim_{x\rightarrow \infty }\frac{e^{x}}{x}$
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 $\lim_{x\rightarrow 0}\frac{x^{2}}{e^{x}-x}$
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 $\lim_{x\rightarrow 0}x.ln(x)$.
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