IB Applications and Interpretation SL · Differential and integral calculus · Section 1 of 7

Limits and the concept of the derivative

3 worked examples · By Álvaro López

Key idea

A limit describes the value that $f (x)$ approaches as $x$ approaches a number, regardless of what the function is exactly at that number. Limits are used to define the derivative, the tool that measures how a function changes at each point.

The idea of a limit

If $x$ approaches $a$ without actually reaching $a$ , the values of $f (x)$ may approach a number $L$ . That is a limit. What matters is what happens around $a$ , not at $a$ itself: the function may not be defined at $a$ , or may take another value there, and the limit still exists.

One-sided limits

Sometimes the behaviour changes depending on the side from which we approach $a$ . That is why one-sided limits are used:

$lim_{x \to a^{-}} f (x)$ : $x$ approaches $a$ from the left, that is, with $x < a$ .
$lim_{x \to a^{+}} f (x)$ : $x$ approaches $a$ from the right, that is, with $x > a$ .

If both one-sided limits equal $L$ , the limit at $a$ exists and equals $L$ . If they differ, or if either one does not exist, the limit at $a$ does not exist. It is like crossing a bridge: you must arrive at the same point from both sides.

Calculating limits with the GDC

There are two ways to estimate a limit: looking at the graph of the function, and looking at a table of values. In the table you choose values of $x$ that get closer and closer to $a$ , some smaller and some larger, and you observe the number to which the values of $f (x)$ approach.

The derivative as a limit

What is the gradient of a curve at a point? A curve does not have a fixed gradient, but it does have a tangent line at each point. To approximate it, we take two points on the graph, $P (a, f (a))$ and $Q (a + h, f (a + h))$ . The line joining them is a secant, and its gradient is

$m_{P Q} = \frac{f ( a + h ) - f ( a )}{h}$

This quotient is called the difference quotient: how much $y$ changes divided by how much $x$ changes. If $Q$ approaches $P$ , that is, if $h$ approaches $0$ , the secant approaches the tangent and its gradient tends to the gradient of the tangent.

Increasing curve y = x²/2 + 1 between x = 0 and x = 4, with points P(1; 1.5) and Q(3; 5.5), the secant PQ, the tangent at P (dashed) and the segments h and f(a+h) − f(a)

For $y = f (x)$ , the derivative is also written $\frac{d y}{d x}$ (Leibniz notation), which is read "the derivative of $y$ with respect to $x$ ".

Worked examples

Example 1 Basic

One-sided limits of a piecewise function

Let $f (x) = {x + 3 2 x if x < 2 if x \geq 2$

a) Calculate $lim_{x \to 2^{-}} f (x)$ .

b) Calculate $lim_{x \to 2^{+}} f (x)$ .

c) State whether $lim_{x \to 2} f (x)$ exists.

Solution

a) From the left, $x < 2$ , so we use the piece $x + 3$ : $x \to 2^{-} lim f (x) = 2 + 3 = 5$

b) From the right, $x \geq 2$ , so we use the piece $2 x$ : $x \to 2^{+} lim f (x) = 2 \cdot 2 = 4$

c) The one-sided limits are different ( $5 \neq = 4$ ), so $lim_{x \to 2} f (x)$ does not exist. Although $f (2) = 4$ , the value of the function at the point does not decide the limit: the graph has a jump at $x = 2$ .

Example 2 Medium

A quotient not defined at the point

Let $g (x) = \frac{x ^{2} - 9}{x - 3}$ . The function is not defined at $x = 3$ , because the denominator equals $0$ .

a) Using the GDC, make a table of values of $g (x)$ for $x = 2.9$ , $x = 2.99$ , $x = 3.01$ and $x = 3.1$ . Calculate $g (2.99)$ .

b) Find $lim_{x \to 3} g (x)$ .

Solution

a) Calculate directly: $g (2.99) = \frac{2.9 9 ^{2} - 9}{2.99 - 3} = \frac{8.9401 - 9}{- 0.01} = \frac{- 0.0599}{- 0.01} = 5.99$ In the table, the values are $g (2.9) = 5.9$ , $g (2.99) = 5.99$ , $g (3.01) = 6.01$ and $g (3.1) = 6.1$ . From both sides, the values approach $6$ .

b) For $x \neq = 3$ we have $x^{2} - 9 = (x - 3) (x + 3)$ , so $g (x) = \frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ Therefore $x \to 3 lim g (x) = 3 + 3 = 6$ The table confirms it, and the result does not change because $g (3)$ does not exist.

Example 3 Hard

Derivative as a limit with a parameter

Let $f (x) = x^{2} + k x$ , where $k$ is a real number.

a) Use the definition of the derivative to find $f^{'} (1)$ in terms of $k$ .

b) Given that $f^{'} (1) = 7$ , find $k$ .

c) With the value of $k$ from the previous part, find $f^{'} (x)$ for any $x$ .

Solution

a) Set up the difference quotient at $x = 1$ : $\frac{f ( 1 + h ) - f ( 1 )}{h} = \frac{( 1 + h ) ^{2} + k ( 1 + h ) - ( 1 + k )}{h}$ Expand the numerator: $(1 + h)^{2} = 1 + 2 h + h^{2}$ and $k (1 + h) = k + k h$ , so $(1 + 2 h + h^{2}) + (k + k h) - (1 + k) = 2 h + h^{2} + k h = h (2 + h + k)$ As $h \neq = 0$ in the limit, we can cancel the $h$ : $f^{'} (1) = h \to 0 lim (2 + k + h) = 2 + k$

b) $2 + k = 7$ , so $k = 5$ .

c) With $k = 5$ we calculate the difference quotient at a general $x$ : $\frac{f ( x + h ) - f ( x )}{h} = \frac{2 x h + h ^{2} + 5 h}{h} = 2 x + h + 5$ Taking the limit as $h \to 0$ , $f^{'} (x) = 2 x + 5$ . Check: at $x = 1$ it gives $7$ , as in part b).

In the exam: write the limit after cancelling the $h$ . Substituting $h = 0$ directly into the quotient gives $\frac{0}{0}$ , which is of no use; you must cancel the $h$ first.

Common mistakes

Summary

The limit $lim_{x \to a} f (x)$ describes the value that $f (x)$ approaches as $x$ approaches $a$ , and does not depend on $f (a)$ .
The limit exists only if the one-sided limits, from the left and from the right, are equal.
The derivative is $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ , the limit of the difference quotient.
Using the GDC you can estimate limits with a table and with the graph, but the result is written and justified by hand.

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