5.1 Limits and the derivativeIB Maths: Applications and Interpretation HL: Revision notes

A limit describes the value that a function $f(x)$ gets closer and closer to as $x$ gets closer and closer to a number $a$. We write $\lim_{x\to a}f(x)=L$.

• The limit depends on the values of $f(x)$ near $x=a$, not necessarily at $x=a$. The function may not even be defined at $a$.
• For the limit to exist, $f(x)$ must approach the same number $L$ from the left ($x<a$) and from the right ($x>a$). Example: $f(x)=\frac{\sqrt{x+9}-3}{x}$ is not defined at $x=0$, but $f(-0.1)=0.16713$, $f(-0.01)=0.16671$, $f(0.01)=0.16662$ and $f(0.1)=0.16621$. The values approach $0.167$ from both sides, so $\lim_{x\to0}f(x)\approx0.167$. At this level you estimate limits from a table or a graph; formal algebraic methods for calculating limits are not required.

To estimate $\lim_{x\to a}f(x)$ from a table:

• calculate $f(x)$ for values of $x$ just below $a$ and just above $a$, getting closer each time (for example $a\pm0.1$, $a\pm0.01$, $a\pm0.001$);
• check that the values from both sides settle towards the same number;
• give the estimate to the accuracy asked for, often $3$ significant figures. Use the table function of your GDC. For $f(x)=\frac{\sqrt{x+9}-3}{x}$, $f(0.0001)=0.166666\ldots$ and $f(-0.0001)=0.166667\ldots$, so the limit is $0.167$ to $3$ s.f. On a graph, trace the curve on both sides of $x=a$ and read the $y$-value the curve heads for. An open circle on the graph shows that the point at $x=a$ is missing, but the limit can still exist.

The gradient of a straight line is constant, but a curve's gradient changes. The gradient of a curve at a point is the gradient of the tangent there. A chord joins two points on the curve. Its gradient between $x=a$ and $x=a+h$ is the average rate of change $\frac{f(a+h)-f(a)}{h}$. As $h$ gets smaller, the chord gets closer to the tangent, so the gradient of the curve at $x=a$ is the limit of the chord gradients as $h\to0$. Example: $h(t)=20t-5t^2$. From $t=1$ the chord gradients are $9.5$ ($h=0.1$), $9.95$ ($h=0.01$) and $9.995$ ($h=0.001$). They approach $10$, so the gradient at $t=1$ is $10$ m s$^{-1}$. This idea is only informal here: you estimate the limit from values; you do not need to prove it.

The derivative of $f$ is the function that gives the gradient of the curve $y=f(x)$ at each value of $x$. It is also called the gradient function. There are several notations for the first derivative:

• $f'(x)$ for a function $f$;
• $\frac{dy}{dx}$ for $y$ in terms of $x$;
• $\frac{dV}{dr}$, $\frac{ds}{dt}$ and so on, naming the variables in context ($V$ against $r$, or $s$ against $t$). The value $f'(a)$, or $\frac{dy}{dx}$ at $x=a$, is the gradient of the tangent at $x=a$. For $h(t)=20t-5t^2$, $h'(1)=10$. Note that $f(a)$ and $f'(a)$ are different things: $f(a)$ is a height on the curve and $f'(a)$ is the gradient there. If $f'(a)>0$ the curve is rising at $x=a$, if $f'(a)<0$ it is falling.

The derivative $\frac{dy}{dx}$ is the rate of change of $y$ with respect to $x$. Its units are the units of $y$ per unit of $x$.

• $\frac{dV}{dt}$ with $V$ in litres and $t$ in minutes is in litres per minute (L min$^{-1}$).
• $\frac{ds}{dt}$ with $s$ in metres and $t$ in seconds is in metres per second (m s$^{-1}$): the velocity. A positive value means the quantity is increasing at that instant, a negative value that it is decreasing. Example: when $t=5$, $\frac{dV}{dt}=12$ means the volume is increasing at $12$ litres per minute at that instant. It does not mean that the volume is $12$ litres. For a small change in $x$ you can estimate the change in $y$: change in $y\approx\frac{dy}{dx}\times$ change in $x$. With $V(5)=60$, $V(5.5)\approx60+12\times0.5=66$ litres. This is an estimate, because the rate itself changes as $x$ changes.