L'Hôpital's rule & limits (HL) – IB Maths exam questions with mark scheme

1CoreNon-calculatorHL only4 marks

Use l'Hôpital's rule to evaluate $\displaystyle\lim_{x\to0}\dfrac{e^{3x}-1}{\sin2x}$.

Both numerator and denominator $\to0$, so form $\tfrac00$ and the rule appliesB1
Differentiates numerator and denominator separatelyM1
$\dfrac{3e^{3x}}{2\cos2x}$A1
Limit $=\dfrac32$A1
Answer: $\dfrac32$
Examiner tip: State the $\frac00$ form before differentiating; the rule is only valid for an indeterminate form.

2StandardNon-calculatorHL only4 marks

Find $\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x^2}$.

Notes form $\tfrac00$ and differentiates numerator and denominator separatelyM1
$\dfrac{\sin x}{2x}$A1
Notes this is still $\tfrac00$ and applies the rule again (accept use of $\lim\frac{\sin x}{x}=1$ here)M1
$\dfrac{\cos x}{2}\to\dfrac12$. Alternative: Maclaurin $1-\cos x=\dfrac{x^2}{2}-\dfrac{x^4}{24}+\dots$ (M1A1), divides by $x^2$ to get $\dfrac12-\dfrac{x^2}{24}+\dots$ (M1), limit $\dfrac12$ (A1)A1
Answer: $\dfrac12$
Examiner tip: Check the form after every differentiation; do not stop at $\frac{\sin x}{2x}$.

3StandardNon-calculatorHL only4 marks

Use l'Hôpital's rule to evaluate $\displaystyle\lim_{x\to\infty}\dfrac{x^2}{e^x}$.

Notes form $\tfrac{\infty}{\infty}$ and differentiates numerator and denominator separatelyM1
$\dfrac{2x}{e^x}$A1
Notes this is still $\tfrac{\infty}{\infty}$ and applies the rule againM1
$\dfrac{2}{e^x}\to0$, so limit $=0$A1
Answer: $0$
Examiner tip: Say why the rule applies each time; 'exponentials beat powers' alone earns no marks when l'Hôpital is requested.

4HardNon-calculatorHL only5 marks

Show that $\displaystyle\lim_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac16$.

Notes form $\tfrac00$ and differentiates numerator and denominator separatelyM1
$\dfrac{1-\cos x}{3x^2}$A1
Notes still $\tfrac00$ and applies the rule again (at least once more)M1
$\dfrac{\sin x}{6x}$A1
$\dfrac{\cos x}{6}$ (or uses $\lim\frac{\sin x}{x}=1$ on $\frac{\sin x}{6x}$), giving $\dfrac16$ AG. Alternative (Maclaurin): $\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dots$ substituted (M1), $x-\sin x=\dfrac{x^3}{6}-\dfrac{x^5}{120}+\dots$ (A1), divides by $x^3$ (M1), $\dfrac16-\dfrac{x^2}{120}+\dots$ (A1), higher terms $\to0$ so limit $\dfrac16$ AG (A1)A1
Answer: $\dfrac16$ (shown)
Examiner tip: It is a 'show that': write every stage, and check the form is still $\frac00$ before each new application of the rule.

5StandardNon-calculatorHL only5 marks

By first writing $x\ln x$ as a quotient, use l'Hôpital's rule to evaluate $\displaystyle\lim_{x\to0^+}x\ln x$.

Rewrites as a quotient, e.g. $\dfrac{\ln x}{1/x}$ (or $\dfrac{\ln x}{x^{-1}}$)M1
Notes form $\tfrac{-\infty}{\infty}$ as $x\to0^+$, so l'Hôpital's rule applies (accept $\tfrac{\infty}{\infty}$ or "indeterminate form")R1
Differentiates: $\dfrac{1/x}{-1/x^2}$M1
$=-x$ (accept unsimplified equivalent, e.g. $\dfrac{-x^2}{x}$)A1
Limit $=0$A1
Answer: $0$
Examiner tip: Product forms $0\times\infty$ must be rewritten as a quotient before the rule can be used; put the logarithm on top.

6StretchNon-calculatorHL only6 marks

By first writing $\dfrac1x-\dfrac1{\sin x}$ as a single fraction, use l'Hôpital's rule to find $\displaystyle\lim_{x\to0}\left(\dfrac1x-\dfrac1{\sin x}\right)$.

Attempts to combine into one fractionM1
$\dfrac{\sin x-x}{x\sin x}$, form $\tfrac00$A1
Differentiates numerator and denominator separately (product rule on the denominator)M1
$\dfrac{\cos x-1}{\sin x+x\cos x}$A1
Notes still $\tfrac00$ and applies the rule again (differentiates numerator and denominator again)M1
$\dfrac{-\sin x}{2\cos x-x\sin x}\to\dfrac02=0$, so limit $=0$A1
Answer: $0$
Examiner tip: The form $\infty-\infty$ is not valid for the rule; combine the fractions first.

7StretchNon-calculatorHL only5 marks

By considering $\ln y$, where $y=(1+2x)^{1/x}$, use l'Hôpital's rule to find $\displaystyle\lim_{x\to0}(1+2x)^{1/x}$.

Takes logs: $\ln y=\dfrac{\ln(1+2x)}{x}$M1
Notes form $\tfrac00$, so l'Hôpital's rule appliesR1
Differentiates: $\dfrac{2/(1+2x)}{1}$M1
$\ln y\to2$A1
$y\to e^2$, so limit $=e^2$A1
Answer: $e^2$
Examiner tip: Finish by exponentiating: the limit of $\ln y$ is $2$, but the limit asked for is $e^2$.

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