Limits, continuity and the derivative from first principles: practice worksheet

1. Question 1warm-up · no calculator

   The function $f$ is defined by $$f(x)=\begin{cases}3-x,&x\leq 2,\\x-1,&x>2.\end{cases}$$

   (a)Find $\displaystyle\lim_{x\to 2}f(x)$.[1]

   Markscheme

   $\displaystyle\lim_{x\to 2}f(x)=1$.

   A1

   Answer: $1$

   (b)Explain why $f$ is continuous but not differentiable at $x=2$.[2]

   Markscheme

   The limit equals $f(2)=1$, establishing continuity.	R1
   The left-hand gradient is $-1$ and the right-hand gradient is $1$; their inequality establishes that $f$ is not differentiable at $x=2$.	R1

   Accept an explanation identifying a corner and its two different gradients.

   Answer: $\displaystyle\lim_{x\to 2}f(x)=f(2)=1$, so $f$ is continuous; the left-hand and right-hand gradients are $-1$ and $1$, so $f$ is not differentiable there.

2. Question 2warm-up · no calculator

   Let $f(x)=3x^2-4x+7$, where $x\in\mathbb{R}$.

   (a)Find $f'(x)$ from first principles.[4]

   Markscheme

   Form $\displaystyle f'(x)=\lim_{h\to 0}\frac{3(x+h)^2-4(x+h)+7-(3x^2-4x+7)}{h}$.	M1
   Expand and collect the numerator to obtain $6xh+3h^2-4h$.	A1
   Simplify the difference quotient to $6x+3h-4$ for $h\neq 0$.	A1
   Take the limit explicitly: $\displaystyle f'(x)=\lim_{h\to 0}(6x+3h-4)=6x-4$.	A1

   A derivative obtained solely by a differentiation rule earns no marks.

   Answer: $f'(x)=6x-4$

3. Question 3exam-style · no calculator

   The curve $C$ has equation $y=f(x)$, where $\displaystyle f(x)=\frac{5}{x+2}$ and $x\neq -2$.

   (a)Show from first principles that $\displaystyle f'(x)=-\frac{5}{(x+2)^2}$.[4]

   Markscheme

   Form $\displaystyle f'(x)=\lim_{h\to 0}\frac{\frac{5}{x+h+2}-\frac{5}{x+2}}{h}$.	M1
   Combine the fractions to obtain $\displaystyle\frac{-5h}{h(x+h+2)(x+2)}$.	A1
   Cancel $h$ to obtain $\displaystyle-\frac{5}{(x+h+2)(x+2)}$.	A1
   Take the limit explicitly to obtain $\displaystyle-\frac{5}{(x+2)^2}$.	A1
   $\displaystyle f'(x)=-\frac{5}{(x+2)^2}$.	AG

   The difference quotient is considered for $h\neq 0$ and $x+h\neq -2$.

   Answer: $\displaystyle f'(x)=-\frac{5}{(x+2)^2}$

   (b)Hence find the equation of the tangent to $C$ at $x=-1$.[2]

   Markscheme

   Obtain the tangent gradient $f'(-1)=-5$.	A1
   Use $f(-1)=5$ to obtain $y-5=-5(x+1)$, or $y=-5x$.	A1

   Accept equivalent equations. Allow FT for the tangent equation using their finite gradient from part (a) and the correct point $(-1,5)$.

   Answer: $y=-5x$

4. Question 4exam-style · no calculator

   Let $g(x)=e^{2x}-1-2x$.

   (a)Write down $g''(x)$ and $g'''(x)$.[2]

   Markscheme

   $g''(x)=4e^{2x}$.	A1
   $g'''(x)=8e^{2x}$.	A1

   Accept equivalent notation for the second and third derivatives.

   Answer: $g''(x)=4e^{2x}$ and $g'''(x)=8e^{2x}$.

   (b)Use l’Hôpital’s rule to determine $\displaystyle\lim_{x\to 0}\frac{g(x)}{x^2}$. State the indeterminate form before each application.[5]

   Markscheme

   State that the original quotient has the indeterminate form $0/0$.	R1
   Obtain $\displaystyle\lim_{x\to 0}\frac{2e^{2x}-2}{2x}$.	A1
   State that the new quotient still has the indeterminate form $0/0$.	R1
   Apply l’Hôpital’s rule again to obtain $\displaystyle\lim_{x\to 0}\frac{4e^{2x}}{2}$.	M1
   Evaluate the limit as $2$.	A1

   Award differentiation marks even if the limit symbol is omitted in intermediate working. The prescribed method must be used.

   Answer: $2$

5. Question 5exam-style · no calculator

   Let $a$ be a positive real number, and define $$L(a)=\lim_{x\to 0}\frac{\sin(ax)-ax}{x-\sin x}.$$ Angles are measured in radians.

   (a)By applying l’Hôpital’s rule, show that $L(a)=-a^3$. State the indeterminate form before each application.[7]

   Markscheme

   State that the original quotient has the indeterminate form $0/0$.	R1
   Obtain $\displaystyle\lim_{x\to 0}\frac{a\cos(ax)-a}{1-\cos x}$.	A1
   State that this quotient has the indeterminate form $0/0$.	R1
   Apply l’Hôpital’s rule again to obtain $\displaystyle\lim_{x\to 0}\frac{-a^2\sin(ax)}{\sin x}$.	M1
   State that this quotient still has the indeterminate form $0/0$.	R1
   Apply l’Hôpital’s rule a third time to obtain $\displaystyle\lim_{x\to 0}\frac{-a^3\cos(ax)}{\cos x}$.	M1
   Substitute $x=0$ to obtain $-a^3$.	A1
   $L(a)=-a^3$.	AG

   The final quotient is no longer indeterminate, so no further application is required.

   Answer: $L(a)=-a^3$

   (b)Hence determine $a$ when $L(a)=-8$.[2]

   Markscheme

   Set $-a^3=-8$, giving $a^3=8$.	M1
   $a=2$.	A1

   Answer: $a=2$

6. Question 6challenge · no calculator

   Let $p>0$. For $x>-\frac{1}{p}$, the function $F$ is defined by $$F(x)=\begin{cases}\dfrac{\ln(1+px)}{x},&x\neq 0,\\c,&x=0.\end{cases}$$

   (a)Use l’Hôpital’s rule to determine $c$, in terms of $p$, so that $F$ is continuous at $x=0$.[3]

   Markscheme

   Identify the indeterminate form $0/0$ in $\displaystyle\lim_{x\to 0}\frac{\ln(1+px)}{x}$.	R1
   Apply l’Hôpital’s rule to obtain $\displaystyle\lim_{x\to 0}\frac{p}{1+px}$.	M1
   The limit is $p$, so continuity requires $c=p$.	A1

   Use $c=p$ in the remaining parts.

   Answer: $c=p$

   (b)Using the definition of the derivative from first principles and l’Hôpital’s rule, show that $\displaystyle F'(0)=-\frac{p^2}{2}$.[6]

   Markscheme

   Form $\displaystyle F'(0)=\lim_{h\to 0}\frac{F(h)-F(0)}{h}=\lim_{h\to 0}\frac{\ln(1+ph)-ph}{h^2}$.	M1
   State that the resulting quotient has the indeterminate form $0/0$.	R1
   Obtain $\displaystyle\lim_{h\to 0}\frac{\frac{p}{1+ph}-p}{2h}$.	A1
   State that this quotient still has the indeterminate form $0/0$.	R1
   Apply l’Hôpital’s rule again to obtain $\displaystyle\lim_{h\to 0}\frac{-p^2/(1+ph)^2}{2}$.	M1
   Evaluate the limit as $\displaystyle-\frac{p^2}{2}$.	A1
   $\displaystyle F'(0)=-\frac{p^2}{2}$.	AG

   Accept algebraic simplification of the first differentiated quotient before taking its limit. The first-principles difference quotient must be shown.

   Answer: $\displaystyle F'(0)=-\frac{p^2}{2}$

   (c)You may use the Maclaurin expansion $$\ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\frac{u^4}{4}+\cdots,$$ valid for $|u|<1$. Use this expansion to find $\displaystyle\lim_{x\to 0}\frac{F(x)-p+\frac{p^2}{2}x}{x^2}$.[2]

   Markscheme

   Obtain $\displaystyle F(x)=p-\frac{p^2}{2}x+\frac{p^3}{3}x^2-\frac{p^4}{4}x^3+\cdots$ near $x=0$.	A1
   Cancel the constant and linear terms, divide by $x^2$, and obtain the limit $\displaystyle\frac{p^3}{3}$.	A1

   It is sufficient to show the expansion through the quadratic term with an indication that further terms follow.

   Answer: $\displaystyle\frac{p^3}{3}$

   (d)The tangent to the curve $y=F(x)$ at $x=0$ passes through $(1,-4)$. Determine $p$ and find the equation of this tangent.[3]

   Markscheme

   Use the tangent equation $\displaystyle y=p-\frac{p^2}{2}x$ and the given point to obtain $\displaystyle-4=p-\frac{p^2}{2}$.	M1
   Solve $(p-4)(p+2)=0$ and select $p=4$ since $p>0$.	A1
   Obtain the tangent equation $y=4-8x$.	A1

   Accept equivalent tangent equations. Allow FT for the final equation using their positive value of $p$ and their gradient from part (b).

   Answer: $p=4$ and $y=4-8x$.

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: Calculus practice.