Write (uv)'' and (uv)'''.

u''v+2u'v'+uv'' and u'''v+3u''v'+3u'v''+uv'''.

How do you choose u in Leibnitz's theorem?

Take u as the polynomial factor, so higher derivatives of u are 0 and few terms remain.

\dfrac{d^n}{dx^n}\left(x^2e^{3x}\right) for n\ge2?

e^{3x}\left(3^nx^2+2n3^{n-1}x+n(n-1)3^{n-2}\right).

State L'Hospital's rule.

If \frac{f}{g} is \frac00 or \frac\infty\infty as x\to a, then \lim\frac fg=\lim\frac{f'}{g'}, provided this limit exists.

What must you check before using L'Hospital's rule?

The form is \frac00 or \frac\infty\infty; otherwise the rule is not valid.

Does L'Hospital's rule use the quotient rule?

No. Differentiate the numerator and the denominator separately.

What if the new quotient is still indeterminate?

Apply the rule again, checking the form each time.

How is \lim{x\to\infty}\left(1+\frac ax\right)^x handled?

Take \ln: \frac{\ln(1+a/x)}{1/x} is \frac00; the rule gives a, so the limit is e^a.

Why can't the rule be used on \lim{x\to0}\frac{1+\cos x}{x^2}?

The numerator tends to 2, not 0; the quotient tends to +\infty.

Leibnitz's theorem and L'Hospital's ruleEdexcel A-Level Further Maths: Flashcards

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State Leibnitz's theorem.

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State Leibnitz's theorem.: $(uv)^{(n)}=\sum_{r=0}^n\binom nr u^{(r)}v^{(n-r)}$ .
Write $(uv)''$ and $(uv)'''$ .: $u''v+2u'v'+uv''$ and $u'''v+3u''v'+3u'v''+uv'''$ .
How do you choose $u$ in Leibnitz's theorem?: Take $u$ as the polynomial factor, so higher derivatives of $u$ are $0$ and few terms remain.
$\dfrac{\mathrm{d}^n}{\mathrm{d}x^n}\left(x^2\mathrm{e}^{3x}\right)$ for $n\ge2$ ?: $\mathrm{e}^{3x}\left(3^nx^2+2n3^{n-1}x+n(n-1)3^{n-2}\right)$ .
State L'Hospital's rule.: If $\frac{f}{g}$ is $\frac00$ or $\frac\infty\infty$ as $x\to a$ , then $\lim\frac fg=\lim\frac{f'}{g'}$ , provided this limit exists.
What must you check before using L'Hospital's rule?: The form is $\frac00$ or $\frac\infty\infty$ ; otherwise the rule is not valid.
Does L'Hospital's rule use the quotient rule?: No. Differentiate the numerator and the denominator separately.
What if the new quotient is still indeterminate?: Apply the rule again, checking the form each time.
$\displaystyle\lim_{x\to0}\frac{\sin3x}{\sin5x}$ ?: $\frac35$ .
$\displaystyle\lim_{x\to0}\frac{2\sin x-\sin2x}{x-\sin x}$ ?: $6$ (three applications).
$\displaystyle\lim_{x\to\infty}\left(1+\frac ax\right)^x$ ?: $\mathrm{e}^a$ .
How is $\lim_{x\to\infty}\left(1+\frac ax\right)^x$ handled?: Take $\ln$ : $\frac{\ln(1+a/x)}{1/x}$ is $\frac00$ ; the rule gives $a$ , so the limit is $\mathrm{e}^a$ .
$\displaystyle\lim_{x\to\infty}\frac{\mathrm{e}^x}{x^2}$ ?: $\infty$ (apply the rule twice).
Why can't the rule be used on $\lim_{x\to0}\frac{1+\cos x}{x^2}$ ?: The numerator tends to $2$ , not $0$ ; the quotient tends to $+\infty$ .

Exam questions on Leibnitz's theorem and L'Hospital's rule

Let $y=x^2\mathrm{e}^{3x}$ .
Use Leibnitz's theorem to find the value of $\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4}$ at $x=0$ .2 marks
L'Hospital's rule states that if $f(x)$ and $g(x)$ both tend to $0$ , or both tend to $\pm\infty$ , as $x\to a$ , then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ , provided the second limit exists.
A student evaluates $\displaystyle\lim_{x\to0}\frac{1+\cos x}{x^2}$ as follows: “This is $\frac00$ . Differentiating gives $\frac{-\sin x}{2x}$ , which is again $\frac00$ . Differentiating again gives $\frac{-\cos x}{2}\to-\frac12$ .” Explain the error in the student's working and describe the true behaviour of the quotient as $x\to0$ .2 marks
Let $f(x)=\dfrac{2\sin x-\sin2x}{x-\sin x}$ for $x\neq0$ .
Show that applying L'Hospital's rule once to $\displaystyle\lim_{x\to0}f(x)$ gives an expression that is again of the form $\frac00$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).