Continuity, first principles and limits (l'Hôpital's rule) – 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.

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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 $e^2$.

8StandardNon-calculatorHL only6 marks

(a) Using differentiation from first principles, show that if $f(x)=x^2+3x$ then $f'(x)=2x+3$. [4]

(b) Find $\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}$. [2]

Uses $f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ with $f(x+h)=(x+h)^2+3(x+h)$ substitutedM1
Numerator simplified correctly: $f(x+h)-f(x)=2xh+h^2+3h$A1
Divides by $h$: $\displaystyle\lim_{h\to0}(2x+h+3)$ (limit notation required on this line, or a clear statement that $h\to0$ is still to be taken)A1
Correct limiting step: as $h\to0$, $2x+h+3\to2x+3$, so $f'(x)=2x+3$ AG. Award only if the limit is taken after cancelling $h$; do not award if $h$ is set to $0$ before dividing.A1
Factorises numerator: $\dfrac{(x-2)(x+2)}{x-2}=x+2$ for $x\neq2$ (accept L'Hôpital's rule: $\displaystyle\lim_{x\to2}\dfrac{2x}{1}$)M1
Limit $=4$A1
Answer: (a) shown; (b) $4$
Examiner tip: Write $\lim_{h\to0}$ on every line until you actually let $h\to0$; examiners penalise dropped limit notation.

9StandardNon-calculatorHL only9 marks

The function $f$ is defined by $$f(x)=\begin{cases}x^2+ax, & 01\end{cases}$$ where $a,b\in\mathbb{R}$.

(a) Given that $f$ is continuous at $x=1$, write down an equation connecting $a$ and $b$. [2]

(b) Given also that $f$ is differentiable at $x=1$, find the value of $a$ and the value of $b$. [3]

(c) Using differentiation from first principles, show that the derivative of $\sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$ for $x>0$. [4]

Equates left and right values at $x=1$: $1^2+a(1)=b\sqrt1$ (accept $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)$ with both sides evaluated)M1
$1+a=b$ (or equivalent)A1
Differentiates both pieces and equates at $x=1$: $2x+a$ and $\dfrac{b}{2\sqrt x}$, giving $2+a=\dfrac b2$ (accept unsimplified)M1
Solves simultaneously with $1+a=b$, e.g. $2+a=\dfrac{1+a}{2}$ (follow through their equation from (a))M1
$a=-3,\ b=-2$ (both required)A1
Uses $\displaystyle\lim_{h\to0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$M1
Multiplies numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$M1
Obtains $\displaystyle\lim_{h\to0}\dfrac{h}{h\left(\sqrt{x+h}+\sqrt x\right)}=\lim_{h\to0}\dfrac{1}{\sqrt{x+h}+\sqrt x}$ (limit notation needed until $h\to0$ is taken)A1
Lets $h\to0$ to get $\dfrac{1}{2\sqrt x}$ AG. Do not award if $h$ is set to $0$ before cancelling, or if l'Hôpital's rule or a standard derivative is used (not first principles).A1
Answer: (a) $b=1+a$; (b) $a=-3,\ b=-2$; (c) shown
Examiner tip: Continuity alone is not enough for differentiability: you must also match the gradients from the left and right at the join.

10HardNon-calculatorHL only7 marks

(a) Using the Maclaurin series for $e^x$ and $\cos x$, show that the Maclaurin series for $e^x\cos x$ up to and including the term in $x^3$ is $1+x-\dfrac{x^3}{3}$. [3]

(b) Hence find $\displaystyle\lim_{x\to0}\dfrac{e^x\cos x-1-x}{x^3}$. [2]

(c) A student says that l'Hôpital's rule gives $\displaystyle\lim_{x\to0}\dfrac{e^x+\sin x}{x+1}=\lim_{x\to0}\dfrac{e^x+\cos x}{1}=2$. Explain why the student is wrong and state the correct value of the limit. [2]

Uses $e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dots$ and $\cos x=1-\dfrac{x^2}{2}+\dots$ and multipliesM1
At least two correct product terms seen, e.g. $\dfrac{x^2}{2}\cdot1$, $1\cdot\left(-\dfrac{x^2}{2}\right)$, $\dfrac{x^3}{6}\cdot1$, $x\cdot\left(-\dfrac{x^2}{2}\right)$A1
Collects terms: $x^2$ terms cancel ($\dfrac12-\dfrac12=0$) and $x^3$ terms give $\dfrac16-\dfrac12=-\dfrac13$, so $1+x-\dfrac{x^3}{3}$ AGA1
Substitutes series: $\dfrac{-\frac{x^3}{3}+(\text{terms in }x^4\text{ and above})}{x^3}$ (accept $-\dfrac13+O(x)$ or $-\dfrac13+\dots$)M1
Limit $=-\dfrac13$ (the question says 'hence', so a method not using the series from (a), e.g. l'Hôpital's rule, earns no marks)A1
At $x=0$ the expression is $\dfrac{1+0}{0+1}=1$, which is not an indeterminate form $\dfrac00$ or $\dfrac{\infty}{\infty}$, so l'Hôpital's rule does not applyR1
Correct limit $=1$ (by direct substitution, since the function is continuous at $x=0$)A1
Answer: (a) shown; (b) $-\dfrac13$; (c) not an indeterminate form; limit $=1$
Examiner tip: Always check you have $\frac00$ or $\frac{\infty}{\infty}$ before using l'Hôpital; if direct substitution works, the limit is just that value.

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