A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther CalculusMethods in CalculusL'Hospital's Rule

L'Hospital's Rule (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 6 January 2026

L'Hospital's rule

What is L'Hospital's rule?

L'Hospital's rule says that $\lim_{x \to a} (\frac{f (x)}{g (x)})$ , i.e. the limit of the expression $\frac{f (x)}{g (x)}$ as $x$ tends towards $a$ ,
- where either
  - $f (a) = 0$ and $g (a) = 0$
  - or $f (a) = \pm \infty$ and $g (a) = \pm \infty$
- is equal to the limit of the derivatives of the numerator and denominator
  - $\lim_{x \to a} (\frac{f (x)}{g (x)}) = \lim_{x \to a} (\frac{f' (x)}{g' (x)})$
- assuming the limit exists

Examiner Tips and Tricks

You must learn L'Hospital's rule, as it is not given in the formulae booklet!

What is the indeterminate form of a limit?

The indeterminate form of a limit is a representation of the nature / type of limit
- although it does not help you actually calculate the limit
  - e.g. $\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $0 \times \infty$ , $\infty - \infty$ , $1^{\infty}$ , $0^{0}$ , $\infty^{0}$
L'Hospital's rule is a method to calculate limits for two of the indeterminate cases above:
- The zero over zero case, $\frac{0}{0}$
  - from $\frac{f (a)}{g (a)}$ when $f (a) = 0$ and $g (a) = 0$
- The infinity over infinity case, $\pm \frac{\infty}{\infty}$
  - from $\frac{f (a)}{g (a)}$ when $f (a) = \pm \infty$ and $g (a) = \pm \infty$

How do I use L'Hospital's rule for the zero over zero case?

An example of the zero over zero case, $\frac{0}{0}$ , of L'Hospital's rule is to find $\lim_{x \to π} (\frac{\sin x}{x - π})$
- Let $f (x) = \sin x$ and $g (x) = x - π$
- Check that $f (π)$ and $g (π)$ are both zero
  - $f (π) = \sin π = 0$ and $g (π) = π - π = 0$
  - so the indeterminate form of the limit is $\frac{0}{0}$
- Use differentiation to find $\frac{f' (x)}{g' (x)}$
  - $\frac{\cos x}{1}$
- Now let $x \to π$
  - $\frac{\cos π}{1} = \frac{- 1}{1} = - 1$
- L'Hospital's rule states that this is the limit of the original expression
  - $\lim_{x \to π} (\frac{\sin x}{x - π}) = - 1$

Examiner Tips and Tricks

Limits with trigonometric functions are always assumed to be in radians unless otherwise specified.

Remember that calculus with trigonometric functions is only valid for radians

Worked Example

Use L'Hospital's rule to determine

$\lim_{x \to 0} (\frac{2 x}{- 1 + e^{x}})$

Answer:

Let $f (x) = 2 x$ and $g (x) = - 1 + e^{x}$

Check if L'Hospital's rule can be used

Calculate $f (0)$ and $g (0)$ and check if $\frac{f (0)}{g (0)}$ is $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$

$\begin{array}{rcl} f (0) & = & 2 \times 0 = 0 \\ g (0) & = & - 1 + e^{0} = - 1 + 1 = 0 \end{array}$

$\lim_{x \to 0} (\frac{f (x)}{g (x)})$ is $\frac{0}{0}$ , so L'Hospital's rule can be used

Find $\frac{f' (x)}{g' (x)}$ by differentiating the top and bottom separately

$\frac{f' (x)}{g' (x)} = \frac{2}{e^{x}}$

Find $\lim_{x \to 0} (\frac{f' (x)}{g' (x)})$ by letting $x$ tend toward 0

$\begin{array}{rcl} \lim_{x \to 0} (\frac{2}{e^{x}}) & = & \frac{2}{e^{0}} \\ = & \frac{2}{1} \\ = & 2 \end{array}$

This is equal to the limit of $\frac{f (x)}{g (x)}$ by L'Hospital's rule

$\lim_{x \to 0} (\frac{2 x}{- 1 + e^{x}}) = 2$

How do I use L'Hospital's rule for the infinity over infinity case?

An example of the infinity over infinity case, $\pm \frac{\infty}{\infty}$ , of L'Hospital's rule is to find $\lim_{x \to 1^{+}} (\frac{\ln (x - 1)}{(\frac{1}{x - 1})})$
- $x \to 1^{+}$ means $x \to 1$ from values above 1
  - e.g. 1.1, 1.01, 1.001, etc
  - but not 0.9, 0.99, 0.999...
  - as the graph of $y = \ln (x - 1)$ doesn't exist for $x < 1$
- Let $f (x) = \ln (x - 1)$ and $g (x) = \frac{1}{x - 1}$
- Check that $f (1)$ and $g (1)$ are both infinite
  - $f (1) = \ln 0 = - \infty$ and $g (1) = \frac{1}{0} = \infty$
  - so the indeterminate form of the limit is $\frac{- \infty}{\infty}$
- Use differentiation to find $\frac{f' (x)}{g' (x)}$
  - $\frac{\frac{1}{x - 1}}{- \frac{1}{{(x - 1)}^{2}}}$
- Simplify before applying the limit
  - $\frac{1}{x - 1} \div \frac{- 1}{{(x - 1)}^{2}} = \frac{1}{(x - 1)} \times \frac{(x - 1) (x - 1)}{- 1} = - (x - 1)$
- Now let $x \to 1^{+}$
  - $- (1 - 1) = 0$
- L'Hospital's rule states that this is the limit of the original expression
  - $\lim_{x \to 1^{+}} (\frac{\ln (x - 1)}{(\frac{1}{x - 1})}) = 0$

Can I use L'Hospital's rule more than once?

L'Hospital's rule can be used repeatedly, so long as the conditions are met each time
For example, to find $\lim_{x \to 0} (\frac{x - \sin x}{x^{2}})$
- Let $f (x) = x - \sin x$ and $g (x) = x^{2}$
- Find $f (0)$ and $g (0)$
  - $f (0) = 0 - \sin 0 = 0$ and $g (0) = 0^{2} = 0$
  - so the indeterminate form of $\lim_{x \to 0} (\frac{f (x)}{g (x)})$ is $\frac{0}{0}$
- Use differentiation to find $\frac{f' (x)}{g' (x)}$
  - $\frac{1 - \cos x}{2 x}$
- But $f' (0) = 1 - \cos 0 = 0$ and $g' (0) = 2 \times 0 = 0$
  - so the indeterminate form of $\lim_{x \to 0} (\frac{f' (x)}{g' (x)})$ is $\frac{0}{0}$
- So differentiate again to find $\frac{f'' (x)}{g'' (x)}$
  - $\frac{\sin x}{1}$
- $f'' (0) = \sin 0 = 0$ and $g'' (0) = 1$
  - this is not $\frac{0}{0}$
  - so $\lim_{x \to 0} (\frac{f'' (x)}{g'' (x)})$ is $\frac{0}{1} = 0$
- By repeated use of L'Hospital's rule, this is the limit of the original expression
  - $\lim_{x \to 0} (\frac{x - \sin x}{x^{2}}) = 0$

Examiner Tips and Tricks

Students often incorrectly use the quotient rule to differentiate $\frac{f (x)}{g (x)}$ instead of just differentiating the top and bottom separately, $\frac{f' (x)}{g' (x)}$ .

How do I rearrange expressions for L'Hospital's rule?

If $\lim_{x \to a} (\frac{f (x)}{g (x)})$ is not of the indeterminate form $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ then L'Hospital's rule cannot be used
- however it is sometimes possible to rearrange $\frac{f (x)}{g (x)}$ into these correct forms
If $\lim_{x \to a} (f (x) g (x))$ has the indeterminate form of the product $0 \times \infty$ , try writing $f$ (or $g$ ) as the reciprocal of its reciprocal, $\frac{1}{1 / f}$
- e.g. $\lim_{x \to 0} (x \cot x)$ has the indeterminate form $0 \times \infty$
  - $\lim_{x \to 0} (\frac{x}{\frac{1}{\cot x}}) = \lim_{x \to 0} (\frac{x}{\tan x})$ now has the correct L'Hospital form of $\frac{0}{0}$
- e.g. $\lim_{x \to 0^{+}} (x \ln x)$ has the indeterminate form $0 \times (- \infty)$
  - $\lim_{x \to 0^{+}} (\frac{\ln x}{\frac{1}{x}})$ now has the correct L'Hospital form of $\frac{- \infty}{\infty}$
If $\lim_{x \to a} (f (x) - g (x))$ has the indeterminate form of the difference $\infty - \infty$ , try writing algebraic fractions as one single simplified fraction
- e.g. $\lim_{x \to 0} (\frac{1}{x} - \cot x)$ has the indeterminate form $\infty - \infty$
  - $\lim_{x \to 0} (\frac{1}{x} - \frac{\cos x}{\sin x}) = \lim_{x \to 0} (\frac{\sin x - x \cos x}{x \sin x})$ now has the correct L'Hospital form of $\frac{0}{0}$

Worked Example

Use L'Hospital's rule to determine

$\lim_{x \to \frac{π}{2}} (\frac{{(x - \frac{π}{2})}^{2} \sec x}{\cos x})$

Answer:

Check if L'Hospital's rule can be used for $\frac{f (x)}{g (x)}$

Calculate $f (\frac{π}{2})$ and $g (\frac{π}{2})$ and check if $\frac{f (\frac{π}{2})}{g (\frac{π}{2})}$ is $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$

$\begin{array}{rcl} f (\frac{π}{2}) & = & {(\frac{π}{2} - \frac{π}{2})}^{2} \sec \frac{π}{2} = 0^{2} \times \frac{1}{\cos \frac{π}{2}} = 0 \times \frac{1}{0} = 0 \times \infty \\ g (\frac{π}{2}) & = & \cos \frac{π}{2} = 0 \end{array}$

$\lim_{x \to \frac{π}{2}} (\frac{f (x)}{g (x)})$ is $\frac{0 \times \infty}{0}$ , so L'Hospital's rule cannot be used

Rearrange $\frac{0 \times \infty}{0}$ to $\frac{0}{0}$ by writing $\frac{{(x - \frac{π}{2})}^{2} \sec x}{\cos x}$ as $\frac{{(x - \frac{π}{2})}^{2}}{\cos^{2} x}$

$\lim_{x \to \frac{π}{2}} (\frac{{(x - \frac{π}{2})}^{2}}{\cos^{2} x})$

Check if L'Hospital's rule can now be used for the new $\frac{f (x)}{g (x)}$

$\begin{array}{rcl} f (\frac{π}{2}) & = & {(\frac{π}{2} - \frac{π}{2})}^{2} = 0^{2} = 0 \\ g (\frac{π}{2}) & = & \cos^{2} \frac{π}{2} = 0^{2} = 0 \end{array}$

$\lim_{x \to \frac{π}{2}} (\frac{f (x)}{g (x)})$ is $\frac{0}{0}$ , so L'Hospital's rule can be used

Find $\frac{f' (x)}{g' (x)}$ by differentiating the top and bottom separately

$\begin{array}{rcl} \frac{f' (x)}{g' (x)} & = & \frac{2 {(x - \frac{π}{2})}^{1}}{2 {(\cos x)}^{1} (- \sin x)} \\ = & \frac{2 (x - \frac{π}{2})}{- 2 \cos x \sin x} \\ = & \frac{x - \frac{π}{2}}{- \cos x \sin x} \end{array}$

Find $\lim_{x \to \frac{π}{2}} (\frac{f' (x)}{g' (x)})$ by letting $x$ tend toward $\frac{π}{2}$

$\begin{array}{rcl} \lim_{x \to \frac{π}{2}} (\frac{x - \frac{π}{2}}{- \cos x \sin x}) & = & \frac{\frac{π}{2} - \frac{π}{2}}{- \cos \frac{π}{2} \sin \frac{π}{2}} \\ = & \frac{0}{- 0 \times 1} \\ = & \frac{0}{0} \end{array}$

This is $\frac{0}{0}$ so requires a second use of L'Hospital's rule

Find $\frac{f'' (x)}{g'' (x)}$ by differentiating again the top and bottom separately

$\begin{array}{rcl} \frac{f'' (x)}{g'' (x)} & = & \frac{1}{(\sin x) \sin x - \cos x (\cos x)} \\ = & \frac{1}{\sin^{2} x - \cos^{2} x} \end{array}$

Find $\lim_{x \to \frac{π}{2}} (\frac{f'' (x)}{g'' (x)})$ by letting $x$ tend toward $\frac{π}{2}$

$\begin{array}{rcl} \lim_{x \to \frac{π}{2}} (\frac{1}{\sin^{2} x - \cos^{2} x}) & = & \frac{1}{1^{2} - 0^{2}} \\ = & \frac{1}{1} \\ = & 1 \end{array}$

This is equal to the limit of $\frac{{(x - \frac{π}{2})}^{2}}{\cos^{2} x}$ by L'Hospital's rule twice

and $\frac{{(x - \frac{π}{2})}^{2}}{\cos^{2} x}$ is a rearrangement of $\frac{{(x - \frac{π}{2})}^{2} \sec x}{\cos x}$

$\lim_{x \to \frac{π}{2}} (\frac{{(x - \frac{π}{2})}^{2} \sec x}{\cos x}) = 1$

How do I take logarithms for L'Hospital's rule?

If $\lim_{x \to a} (f {(x)}^{g (x)})$ has the indeterminate form of the power $1^{\infty}$ , or $0^{0}$ , or $\infty^{0}$ , etc, try
- taking the logarithm of the function inside the limit, $f {(x)}^{g (x)}$
  - using log laws to write it in the form $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$
- using L'Hospital's rule to find the limit
- then writing the answer to the original limit as $e^{limit}$
e.g. $\lim_{x \to 0} ({(1 + \sin x)}^{\frac{1}{x}})$ has the indeterminate form $1^{\infty}$
- So take logs of the function
  - $\lim_{x \to 0} (\ln ({(1 + \sin x)}^{\frac{1}{x}}))$
- Use log laws to simplify
  - $\lim_{x \to 0} (\frac{1}{x} \ln (1 + \sin x)) = \lim_{x \to 0} (\frac{\ln (1 + \sin x)}{x})$
  - This now has the correct L'Hospital form of $\frac{0}{0}$
- Use L'Hospital's rule to find the limit
  - $\lim_{x \to 0} (\frac{\ln (1 + \sin x)}{x}) = \lim_{x \to 0} (\frac{\frac{\cos x}{1 + \sin x}}{1}) = \frac{\frac{1}{1 + 0}}{1} = 1$
- The final answer is $e^{limit}$
  - $\lim_{x \to 0} ({(1 + \sin x)}^{\frac{1}{x}}) = e^{1} = e$

Worked Example

Use L'Hospital's rule to determine

$\lim_{x \to \infty} ({(1 + \frac{a}{x})}^{x})$

Answer:

Find the indeterminate form of the limit given

$\lim_{x \to \infty} ({(1 + \frac{a}{x})}^{x}) = {(1 + 0)}^{\infty} = 1^{\infty}$

This is not in the form $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ for L'Hospital's rule

Take logs of the function inside the limit

$\ln {(1 + \frac{a}{x})}^{x}$

Use log laws to simplify

$x \ln (1 + \frac{a}{x})$

Check the indeterminate form of this limit

$\lim_{x \to \infty} (x \ln (1 + \frac{a}{x})) = \infty \times \ln 1 = \infty \times 0$

This is still not in the form $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ for L'Hospital's rule

Try writing $x$ as $\frac{1}{(\frac{1}{x})}$

$\lim_{x \to \infty} (\frac{\ln (1 + \frac{a}{x})}{\frac{1}{x}}) = \frac{\ln 1}{0} = \frac{0}{0}$

This is now in the correct form $\frac{0}{0}$ for L'Hospital's rule

Find $\frac{f' (x)}{g' (x)}$ by differentiating the top and bottom separately, then simplify

$\begin{array}{rcl} \frac{f' (x)}{g' (x)} & = & \frac{(\frac{- \frac{a}{x^{2}}}{1 + \frac{a}{x}})}{- \frac{1}{x^{2}}} \\ = & \frac{- \frac{a}{x^{2}}}{1 + \frac{a}{x}} \div \frac{- 1}{x^{2}} \\ = & \frac{- \frac{a}{x^{2}}}{1 + \frac{a}{x}} \times \frac{x^{2}}{- 1} \\ = & \frac{- a}{- 1 - \frac{a}{x}} \\ = & \frac{a}{1 + \frac{a}{x}} \end{array}$

Find $\lim_{x \to \infty} (\frac{f' (x)}{g' (x)})$ by letting $x$ tend toward $\infty$

$\begin{array}{rcl} \lim_{x \to \infty} (\frac{f' (x)}{g' (x)}) & = & \frac{a}{1 + 0} \\ = & a \end{array}$

This is equal to the limit of $\frac{f (x)}{g (x)}$ by L'Hospital's rule

$\lim_{x \to \infty} (\frac{\ln (1 + \frac{a}{x})}{\frac{1}{x}}) = a$

This means the limit of the original function, before taking logarithms, is $e^{a}$

$\lim_{x \to \infty} ({(1 + \frac{a}{x})}^{x}) = e^{a}$

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L'Hospital's Rule (Edexcel A Level Further Maths: Further Pure 1): Revision Note

L'Hospital's rule

What is L'Hospital's rule?

What is the indeterminate form of a limit?

How do I use L'Hospital's rule for the zero over zero case?

How do I use L'Hospital's rule for the infinity over infinity case?

Can I use L'Hospital's rule more than once?

How do I rearrange expressions for L'Hospital's rule?

How do I take logarithms for L'Hospital's rule?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis