A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther CalculusTaylor SeriesLimits using Series

Limits using Series (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 6 January 2026

Limits using series

How do I find limits using Taylor series?

To find the limit of a function as $x$ tends to $a$ , $\lim_{x \to a} f (x)$
- first write $f (x)$ as a Taylor series about $x = a$
- then let $x \to a$ in each term
  - Each ${(x - a)}^{n}$ term becomes zero leaving behind the limit
This method works for combinations of functions
- e.g. $\lim_{x \to a} (\frac{f (x)}{g (x)})$ , $\lim_{x \to a} (f (x) \pm g (x))$ or $\lim_{x \to a} (f (x) g (x))$
  - First write $f (x)$ and $g (x)$ as a Taylor series about $x = a$
  - Simplify the algebra inside the limit
  - Then let $x \to a$ and see what is left behind

Examiner Tips and Tricks

You will be given the Taylor series expansion in the exam question:

$f (x) = f (a) + (x - a) f' (a) + \frac{{(x - a)}^{2}}{2!} f'' (a) + . . . + \frac{{(x - a)}^{r}}{r!} f^{(r)} (a) + . . .$

How do I find limits using Maclaurin series?

In the special case when $a = 0$
- the limit is $\lim_{x \to 0} f (x)$
- you can use Maclaurin series
  - $f (x) = f (0) + x f' (0) + \frac{x^{2}}{2!} f'' (0) + . . . + \frac{x^{r}}{r!} f^{(r)} (0) + . . .$
This is particularly useful for functions inside functions
- e.g. $\lim_{x \to 0} (\frac{\sin (x^{3})}{x^{3}})$
  - use $\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . .$ from the formulae booklet
- substitute in $x^{3}$
  - $\sin (x^{3}) = x^{3} - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - . . .$
- cancel and take the limit
  - $\lim_{x \to 0} (\frac{x^{3} (1 - \frac{x^{6}}{3!} + \frac{x^{12}}{5!} -)}{x^{3}}) = 1 - 0 + 0 - . . . = 1$

Examiner Tips and Tricks

You are given the Maclaurin series formula in the formulae booklet, as well as Maclaurin series expansions for $e^{x}$ , $\ln (1 + x)$ , $\sin x$ , $\cos x$ and $\arctan x$ .

Worked Example

By finding the Taylor series expansion about $x = 1$ of $\ln x$ in ascending powers of $(x - 1)$ , up to and including the term in ${(x - 1)}^{4}$ , find

$\lim_{x \to 1} (\frac{\ln x - (x - 1) + \frac{1}{2} {(x - 1)}^{2}}{{(x - 1)}^{3}})$

Answer:

First find the Taylor series by letting $f (x) = \ln x$ and finding the first four derivatives

$\begin{array}{rcl} f' (x) & = & \frac{1}{x} \\ f'' (x) & = & - \frac{1}{x^{2}} \\ f''' (x) & = & \frac{2}{x^{3}} \\ f^{(4)} (x) & = & - \frac{6}{x^{4}} \end{array}$

Substitute $x = 1$ into $f (x)$ and its derivatives

$\begin{array}{rcl} f (1) & = & \ln 1 = 0 \\ f' (1) & = & \frac{1}{1} = 1 \\ f'' (1) & = & - \frac{1}{1^{2}} = - 1 \\ f''' (1) & = & \frac{2}{1^{3}} = 2 \\ f^{(4)} (1) & = & - \frac{6}{1^{4}} = - 6 \end{array}$

Substitute these values into the Taylor series expansion given by

$f (x) = f (a) + (x - a) f' (a) + \frac{{(x - a)}^{2}}{2!} f'' (a) + . . . + \frac{{(x - a)}^{r}}{r!} f^{(r)} (a) + . . .$

$\begin{array}{rcl} \ln x & = & 0 + (x - 1) \times 1 + \frac{{(x - 1)}^{2}}{2!} \times (- 1) + \frac{{(x - 1)}^{3}}{3!} \times 2 + \frac{{(x - 1)}^{4}}{4!} \times (- 6) + . . . \\ = & (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3} - \frac{1}{4} {(x - 1)}^{4} + . . . \end{array}$

Now substitute this Taylor series into the expression in the question, cancelling any terms

$\frac{\ln x - (x - 1) + \frac{1}{2} {(x - 1)}^{2}}{{(x - 1)}^{3}} = \frac{(x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3} - \frac{1}{4} {(x - 1)}^{4} + . . . - (x - 1) + \frac{1}{2} {(x - 1)}^{2}}{{(x - 1)}^{3}}$

Factorise out ${(x - 1)}^{3}$ from top and bottom and cancel

$\begin{array}{rcl} = & \frac{\frac{1}{3} {(x - 1)}^{3} - \frac{1}{4} {(x - 1)}^{4} + . . .}{{(x - 1)}^{3}} \\ = & \frac{{(x - 1)}^{3} (\frac{1}{3} - \frac{1}{4} (x - 1) + . . .)}{{(x - 1)}^{3}} \\ = & \frac{1}{3} - \frac{1}{4} (x - 1) + . . . \end{array}$

Now take the limit as $x \to 1$ (i.e. all terms involving ${(x - 1)}^{n}$ go to zero)

$\begin{array}{rcl} \lim_{x \to 1} (\frac{\ln x - (x - 1) + \frac{1}{2} {(x - 1)}^{2}}{{(x - 1)}^{3}}) & = & \lim_{x \to 1} (\frac{1}{3} - \frac{1}{4} (x - 1) + . . .) \\ = & \frac{1}{3} + \lim_{x \to 1} (- \frac{1}{4} (x - 1) + . . . .) \\ = & \frac{1}{3} + 0 \\ = & \frac{1}{3} \end{array}$

Write out the final result

$\lim_{x \to 1} (\frac{\ln x - (x - 1) + \frac{1}{2} {(x - 1)}^{2}}{{(x - 1)}^{3}}) = \frac{1}{3}$

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