Limits using Maclaurin series and L'Hopital's ruleAQA A-Level Further Maths: Revision notes

Substituting the limit value sometimes gives $\frac00$ or $\frac{\infty}{\infty}$. These are indeterminate forms: they have no value on their own and the limit may be any number, so another method is needed. Examples: $\lim_{x\to0}\frac{\sin x}{x}$ is $\frac00$ at $x=0$, and $\lim_{x\to\infty}\frac{x^2}{\mathrm{e}^x}$ is $\frac{\infty}{\infty}$. Other forms, such as $0\times\infty$ or $\infty-\infty$, must first be rewritten as a quotient. For example $x\ln x$ as $x\to0^+$ becomes $\frac{\ln x}{1/x}$, which is $\frac{-\infty}{\infty}$. Always substitute first: if the limit is not indeterminate, it is the value you get.

If $\frac{f(x)}{g(x)}$ has the form $\frac00$ or $\frac{\infty}{\infty}$ as $x\to a$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ provided the right-hand limit exists. Differentiate the numerator and denominator separately (this is not the quotient rule). If the result is still indeterminate, apply the rule again. Example: $\lim_{x\to0}\frac{1-\cos2x}{x^2}=\lim\frac{2\sin2x}{2x}$ (still $\frac00$) $=\lim\frac{4\cos2x}{2}=2$. Example, $x\to\infty$: $\lim\frac{x^2}{\mathrm{e}^x}=\lim\frac{2x}{\mathrm{e}^x}=\lim\frac{2}{\mathrm{e}^x}=0$. More generally $\mathrm{e}^x$ grows faster than any power of $x$, so $\frac{x^k}{\mathrm{e}^x}\to0$.

As $x\to0$, replace each function by its Maclaurin series, simplify, and divide by the lowest power of $x$. Terms with higher powers vanish in the limit. Example: $\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}$. Since $\sin x=x-\frac{x^3}{6}+\dots$ and $x\cos x=x-\frac{x^3}{2}+\dots$, the numerator is $\frac{x^3}{3}+\dots$ and the limit is $\frac13$. Example: $\lim_{x\to0}\frac{\mathrm{e}^x-1-x}{x^2}$: the numerator is $\frac{x^2}{2}+\frac{x^3}{6}+\dots$, so the limit is $\frac12$. Take enough terms: the series must be taken far enough that the leading non-zero term of the numerator appears. If it cancels completely, include the next term.

• Maclaurin series suits limits as $x\to0$ involving standard functions ($\mathrm{e}^x,\sin x,\cos x,\ln(1+x)$); it shows how the expression behaves.
• L'Hôpital is quicker when the derivatives simplify, and is the method for limits as $x\to\infty$, where a Maclaurin series is of no use.
• For $\lim_{x\to0}\frac{\sin 3x}{\sin 5x}$: one application gives $\frac{3\cos3x}{5\cos5x}\to\frac35$. Whichever you use, show the indeterminate form, the working at each stage, and the final value. Verifying a result by the second method is good practice when time allows.