Limits and Derivatives

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 12.

How fast are you going, right now?

Your speedometer shows 60 km/h at this very instant. But speed is distance divided by time, and at one instant no time passes and no distance is covered, which would make the speed $0/0$. So how can a speedometer show a number at all? It measures the distance over a tiny stretch of time and shrinks the stretch. The number the readings settle towards is the instantaneous speed. That settling-down idea is a limit, and the instantaneous rate of change it produces is the derivative. This chapter builds both.

The idea in one falling stone

A stone falls so that its distance after $t$ seconds is $s = 4.9t^2$ metres. Take the stretch from $t = 2$ to $t = 2 + h$. The average velocity there is

$$\dfrac{4.9(2+h)^2 - 4.9(2)^2}{h} = 4.9(4 + h) = 19.6 + 4.9h$$

Now shrink $h$, and also look at stretches that end at $t = 2$.

The readings from the right come down and the readings from the left go up, and both close in on one number, $19.6$ m/s. That number is the velocity at $t = 2$. Geometrically, each average velocity is the slope of a chord of the graph, and the chords swing into the tangent at the point as $h \to 0$.

Limits: what the function is heading towards

$\lim_{x \to a} f(x) = l$ means $f(x)$ gets as close to $l$ as we like when $x$ is close to $a$, without caring what happens at $x = a$ itself. You can approach $a$ from the left, giving the left hand limit $\lim_{x \to a^{-}} f(x)$, or from the right, giving the right hand limit $\lim_{x \to a^{+}} f(x)$.

$$\lim_{x \to a} f(x) \text{ exists} \iff \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$

The value of the limit and the value $f(a)$ are independent facts. Three cases keep appearing.

Pause and predict: a function has left hand limit $3$ and right hand limit $3$ at $x = a$, but $f(a) = 5$. What is $\lim_{x \to a} f(x)$?

Check your answer: $3$. The limit only listens to the values near $a$. The odd value $5$ at $a$ itself does not matter.

Beyond the textbook. For the exam: a limit exists when the left and right limits agree. In reality: the hole in the graph of $\dfrac{x^2-4}{x-2}$ at $x = 2$ is called a removable gap, and filling it with the limit value $4$ repairs the function. Mathematicians make "gets as close as we like" exact with the $\varepsilon$ and $\delta$ definition of a limit, which is where the idea comes from in higher mathematics. Why it matters: it explains why $0/0$ is not an answer but a signal to simplify.

Algebra of limits

If $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist, limits behave exactly like arithmetic.

Two consequences matter most. For a polynomial $p$, $\lim_{x \to a} p(x) = p(a)$, so you just substitute. For a rational function $\dfrac{g(x)}{h(x)}$ the answer depends on what substitution gives.

If after cancelling the numerator still vanishes and the denominator does not, the limit is $0$. If the denominator still vanishes and the numerator does not, the limit does not exist.

Worked example: a 0 by 0 form. Evaluate $\lim_{x \to 2} \dfrac{x^3 - 4x^2 + 4x}{x^2 - 4}$.

Common wrong step. Stopping at $0/0$ and writing "undefined" or "does not exist". A $0/0$ form says nothing about the limit. It only tells you to factor.

Standard limit for powers

$$\lim_{x \to a} \dfrac{x^n - a^n}{x - a} = n\,a^{n-1}$$

Why it holds. For a positive integer $n$, $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \dots + a^{n-1})$. Cancel $(x-a)$. The bracket has $n$ terms, and each tends to $a^{n-1}$ as $x \to a$. So the total is $n\,a^{n-1}$. The result also holds for any rational power $n$ when $a > 0$.

Worked example: a ratio of two power limits. Evaluate $\lim_{x \to 1} \dfrac{x^{15} - 1}{x^{10} - 1}$.

Common wrong step. Writing the answer to the standard limit as $n\,a^n$. The exponent drops by one. If $a = 1$ you cannot see the mistake, so test yourself with $a = 2$: $\lim_{x \to 2} \dfrac{x^5 - 32}{x - 2} = 5 \times 2^4 = 80$.

The same limit handles roots when you shift the variable. For $\lim_{x \to 0} \dfrac{\sqrt{1+x} - 1}{x}$, put $y = 1 + x$. Then $y \to 1$, and the limit becomes $\dfrac{y^{1/2} - 1}{y - 1} \to \dfrac{1}{2}\,1^{-1/2} = \dfrac{1}{2}$.

Trigonometric limits

$$\lim_{x \to 0} \dfrac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0, \qquad \lim_{x \to 0} \dfrac{\tan x}{x} = 1$$

Why $\dfrac{\sin x}{x} \to 1$. In a unit circle, for $0 < x < \dfrac{\pi}{2}$, compare the area of a small triangle, a sector and a big triangle. They give $\sin x < x < \tan x$. Divide by $\sin x$ to get $\cos x < \dfrac{\sin x}{x} < 1$. As $x \to 0$, $\cos x \to 1$, so the middle term is sandwiched and also tends to $1$. Because $\dfrac{\sin x}{x}$ is unchanged when $x$ is replaced by $-x$, the left limit is the same.

Why $\dfrac{1 - \cos x}{x} \to 0$. Use $1 - \cos x = 2\sin^2 \dfrac{x}{2}$. Then $\dfrac{1 - \cos x}{x} = \sin\dfrac{x}{2} \cdot \dfrac{\sin(x/2)}{x/2}$, which tends to $0 \times 1 = 0$.

The third follows from $\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x} \to 1 \times 1$.

Pause and predict: does $\dfrac{1 - \cos x}{x}$ tend to $1$ or $0$ as $x \to 0$? Try $x = 0.1$ ($\cos 0.1 \approx 0.995004$).

Check your answer: $0$. At $x = 0.1$ the ratio is about $0.0500$, and at $x = 0.01$ it is about $0.0050$. The top shrinks like $x^2$, faster than the bottom shrinks like $x$.

Worked example: matching the argument. Evaluate $\lim_{x \to 0} \dfrac{\sin 4x}{\sin 2x}$.

Common wrong step. Replacing both sine ratios by $1$ and forgetting the leftover factor $\dfrac{4x}{2x} = 2$, which gives $1$. Whenever the two arguments differ, the leftover factor is what the answer is made of.

Beyond the textbook. For the exam: $\dfrac{\sin x}{x} \to 1$, with $x$ in radians. In reality: if $x$ is measured in degrees, the limit is $\dfrac{\pi}{180} \approx 0.0175$, because $x$ degrees is $\dfrac{\pi x}{180}$ radians. That is why all of calculus uses radians: $\dfrac{d}{dx}(\sin x) = \cos x$ holds only in radians, and in degrees it would carry an ugly factor $\dfrac{\pi}{180}$. Why it matters: a limit or derivative question that mentions degrees is a trap.

The derivative

The step from the falling stone to a general rule is just notation. The derivative of $f$ at $a$ is

$$f'(a) = \lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}$$

provided the limit exists. It is the instantaneous rate of change of $f$ at $a$, and it is the slope of the tangent to $y = f(x)$ at $(a, f(a))$, because $\dfrac{f(a+h) - f(a)}{h}$ is the slope of the chord from $(a, f(a))$ to $(a+h, f(a+h))$. Replace $a$ by $x$ and you get the derivative as a function. This is the first principle definition.

$$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}, \qquad f'(x) \text{ also written } \dfrac{dy}{dx}$$

Back to the stone: $s = 4.9t^2$ gives $\dfrac{ds}{dt} = 9.8t$, so the velocity at $t = 2$ is $19.6$ m/s, as the table suggested.

Worked example: first principle. Find the derivative of $f(x) = \dfrac{1}{x}$ from the definition.

Common wrong step. Writing $\dfrac{1}{x+h} - \dfrac{1}{x} = \dfrac{1}{h}$, treating the subtraction as if denominators could be subtracted. Always put the two fractions over a common denominator first.

Rules of differentiation

For differentiable $u$ and $v$:

The power rule is the standard limit again. Expanding $(x+h)^n$ by the binomial theorem gives $\dfrac{(x+h)^n - x^n}{h} = n x^{n-1} + \binom{n}{2} x^{n-2}h + \dots$, and every term except the first has a factor of $h$ that vanishes. So a polynomial is differentiated term by term: the derivative of $a_n x^n + \dots + a_1 x + a_0$ is $n a_n x^{n-1} + \dots + a_1$.

The trigonometric derivatives come from the same machinery.

Worked example: product rule. Differentiate $y = x^2 \sin x$.

Common wrong step. Multiplying the derivatives, $u'v'$, which gives $2x\cos x$. The product rule has two terms, and each term keeps one original function.

Worked example: quotient rule. Differentiate $y = \dfrac{1 + x}{1 - x}$.

Common wrong step. Reversing the numerator to $uv' - u'v$, which flips the sign and gives $-\dfrac{2}{(1-x)^2}$. The numerator is the derivative of the top times the bottom, minus the top times the derivative of the bottom. Write $u'v - uv'$ before you substitute anything.

Beyond the textbook. For the exam: the derivative is the limit of the difference quotient, and its rules are listed above. In reality: the derivative is the language of change. Velocity is the derivative of position, acceleration is the derivative of velocity, and a business looks at the derivative of cost to see what one more unit will cost. The same limit idea was developed independently by Newton and Leibnitz, who used it to describe motion and the slope of curves. Why it matters: every Class XII topic on rates, maxima and minima, and integration rests on the rules you practise here.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before: Introduction to Three Dimensional Geometry and Trigonometric Functions, whose identities are used in the trig limits.
• Next in Class XI: Statistics moves from change to spread.
• Class XII: Continuity and Differentiability turns the limit ideas here into formal definitions, Application of Derivatives puts the derivative to work on rates and maxima, and Integrals reverses it.