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

Leibnitz's theorem
Using it
L'Hospital's rule

Leibnitz and L'Hospital

products and limits

LeibnitzL'Hospital
Repeating
Other forms
Exam tips

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

  1. Let y=x2e3xy=x^2\mathrm{e}^{3x}.
    Use Leibnitz's theorem to find the value of d4ydx4\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} at x=0x=0.2 marks
  2. L'Hospital's rule states that if f(x)f(x) and g(x)g(x) both tend to 00, or both tend to ±∞\pm\infty, as x→ax\to a, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\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 lim⁡x→01+cos⁡xx2\displaystyle\lim_{x\to0}\frac{1+\cos x}{x^2} as follows: “This is 00\frac00. Differentiating gives −sin⁡x2x\frac{-\sin x}{2x}, which is again 00\frac00. Differentiating again gives −cos⁡x2→−12\frac{-\cos x}{2}\to-\frac12.” Explain the error in the student's working and describe the true behaviour of the quotient as x→0x\to0.2 marks
  3. Let f(x)=2sin⁡x−sin⁡2xx−sin⁡xf(x)=\dfrac{2\sin x-\sin2x}{x-\sin x} for x≠0x\neq0.
    Show that applying L'Hospital's rule once to lim⁡x→0f(x)\displaystyle\lim_{x\to0}f(x) gives an expression that is again of the form 00\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).