Further Pure 1 - Further Calculus - Series Expansions and Limits

We can find certain limits of functions by using their Taylor series. To take an example:

\displaystyle \lim_{x \to 0} \frac {x - \arctan x} {x^3}

We know that the arctangent has series expansion:

\displaystyle \arctan x = x - \frac {x^3} 3 + \frac {x^5} 5 + \ldots

So:

\begin{align*}\frac {x - \arctan x} {x^3} & = \frac {x - \left(x - \frac {x^3} 3 + \frac {x^5} 5 + \ldots\right)} {x^3} \\ & = \frac {\frac {x^3} 3 - \frac {x^5} 5 - \ldots} {x^3} \\ & = \frac 1 3 - \frac {x^2} 5 - \ldots\end{align*}

We see that as x \to 0, the terms after \dfrac 1 3 will all go to zero, (since they are all positive powers of x) so:

\displaystyle \lim_{x \to 0} \frac {x - \arctan x} {x^3} = \frac 1 3

Watch for circularity, if you are trying to find the derivative of a function using a limit, but use the function’s Taylor series to do this, your proof may be circular!

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