Limits and Derivatives

A limit tells us the value a function f(x) gets closer and closer to as the input x gets closer and closer to some fixed number a — without necessarily x ever reaching a, and without f(a) even needing to be defined.

To check whether a limit exists at x = a, we examine the function from both sides of a:

• The left-hand limit (LHL) is the value f(x) approaches as x approaches a through values less than a, written lim_x→a^- f(x).
• The right-hand limit (RHL) is the value f(x) approaches as x approaches a through values greater than a, written lim_x→a⁺ f(x).

The limit lim_x→a f(x) exists only when the LHL and RHL are equal; their common value is then the limit.

Worked example. Consider f(x) = (x² - 1)/(x - 1) for x ≠ 1. Although f(1) is not defined (0/0 form), we simplify: (x²-1)/(x-1) = (x-1)(x+1)/(x-1) = x + 1 for every x ≠ 1. As x approaches 1 from either side, x + 1 approaches 2. So the left-hand limit and right-hand limit both equal 2, hence lim_x→1 (x²-1)/(x-1) = 2, even though the function itself is undefined at x = 1.

Once the algebra of limits is known, limits of polynomial and rational functions can be evaluated directly, without guesswork. If lim_x→a f(x) and lim_x→a g(x) both exist, then:

• lim_x→a [f(x) + g(x)] = lim_x→a f(x) + lim_x→a g(x)
• lim_x→a [f(x) - g(x)] = lim_x→a f(x) - lim_x→a g(x)
• lim_x→a [f(x)·g(x)] = lim_x→a f(x) · lim_x→a g(x)
• lim_x→a [f(x)/g(x)] = lim_x→a f(x) / lim_x→a g(x), provided lim_x→a g(x) ≠ 0

For a polynomial P(x) = a₀ + a₁x + a₂x² + ... + a_nxⁿ, repeated use of these rules gives lim_x→a P(x) = P(a) — simply substitute x = a. For a rational function f(x)/g(x) with g(a) ≠ 0, the limit is f(a)/g(a). If direct substitution gives the indeterminate form 0/0, both numerator and denominator share a factor (x - a); cancel it first, then substitute again.

Worked example. Evaluate lim_x→3 (x⁴ - 81)/(2x² - 5x - 3). Direct substitution gives 0/0, so we factor both parts. Numerator: x⁴ - 81 = (x² - 9)(x² + 9) = (x - 3)(x + 3)(x² + 9). Denominator: 2x² - 5x - 3 = (x - 3)(2x + 1). Cancelling the common factor (x - 3):

lim_x→3 [(x + 3)(x² + 9)]/(2x + 1) = (3 + 3)(9 + 9)/(2·3 + 1) = (6)(18)/7 = 108/7.

Three limits form the backbone of every trigonometric-limit problem at this level, and are stated as standard results:

• lim_x→0 sinx/x = 1 (x measured in radians)
• lim_x→0 (1 - cosx)/x = 0
• lim_x→0 tanx/x = 1, obtained from tanx/x = (sinx/x)·(1/cosx), using cos0 = 1

These give a useful general pattern: for any non-zero constant k, lim_x→0 sin(kx)/(kx) = 1, so lim_x→0 sin(kx)/x = k.

Worked example. Evaluate lim_x→0 sin3x/sin5x. Multiply and divide so each trig ratio matches the standard form:

sin3x/sin5x = {[sin3x/(3x)]·3x} / {[sin5x/(5x)]·5x} = {[sin3x/(3x)] / [sin5x/(5x)]} · (3x/5x).

As x → 0, both sin3x/(3x) and sin5x/(5x) approach 1, and 3x/5x = 3/5 for all x ≠ 0. So the limit equals (1/1)·(3/5) = 3/5.

The derivative of a function measures how fast it is changing — the instantaneous rate of change of y = f(x) with respect to x.

If x changes from x to x + h, the average rate of change of f over this interval is [f(x+h) - f(x)]/h. As h is made smaller and smaller, this average rate approaches the instantaneous rate of change at x, called the derivative of f at x, written f'(x) or dy/dx:

f'(x) = lim_h→0 [f(x+h) - f(x)]/h

Finding a derivative directly from this limit definition — without using any shortcut rule — is called differentiating from first principles.

Worked example (first principles). Find the derivative of f(x) = x² from first principles. f(x+h) - f(x) = (x+h)² - x² = 2xh + h² = h(2x + h). So [f(x+h)-f(x)]/h = 2x + h. Taking the limit as h → 0 gives f'(x) = 2x.

Worked example (rate of change). The distance covered by a particle is s(t) = t² + 3t (t in seconds, s in metres). Its velocity is s'(t), the rate of change of s with respect to t. Using the power rule (x replaced by t), s'(t) = 2t + 3. At t = 2 s, the velocity is s'(2) = 2(2) + 3 = 7 m/s.

Finding every derivative from first principles is slow. A small set of standard rules, each itself provable from first principles, lets us differentiate combinations of functions quickly.

If u = u(x) and v = v(x) are differentiable functions and c is a constant:

• Sum/Difference rule: d/dx(u ± v) = du/dx ± dv/dx
• Constant multiple rule: d/dx(c·u) = c·du/dx
• Product rule (Leibniz rule): d/dx(u·v) = u·(dv/dx) + v·(du/dx)
• Quotient rule: d/dx(u/v) = [v·(du/dx) - u·(dv/dx)]/v², provided v ≠ 0

Worked example (product rule). Differentiate f(x) = x²sinx. Here u = x², v = sinx, so du/dx = 2x and dv/dx = cosx. By the product rule, f'(x) = x²·cosx + sinx·2x = x²cosx + 2x·sinx.

Worked example (quotient rule). Differentiate tanx = sinx/cosx using the quotient rule, with u = sinx, v = cosx: du/dx = cosx, dv/dx = -sinx.

d/dx(tanx) = [cosx·cosx - sinx·(-sinx)]/cos²x = (cos²x + sin²x)/cos²x = 1/cos²x = sec²x, using sin²x + cos²x = 1.

Combining the power rule with the derivatives of the trigonometric functions gives a ready reference for differentiating any polynomial or trigonometric expression in board or entrance exams.

Using the limit result lim_x→a (xⁿ - aⁿ)/(x - a) = n·a^n-1, the derivative of xⁿ follows directly from first principles:

d/dx(xⁿ) = n·x^n-1, for any rational number n.

The derivatives of sinx and cosx also follow from first principles, using compound-angle identities and lim_h→0 sinh/h = 1:

• d/dx(sinx) = cosx
• d/dx(cosx) = -sinx

The remaining trigonometric derivatives follow from these two by the quotient rule (as shown for tanx earlier):

• d/dx(tanx) = sec²x
• d/dx(cotx) = -cosec²x
• d/dx(secx) = secx·tanx
• d/dx(cosecx) = -cosecx·cotx

Worked example. Differentiate f(x) = 3x⁴ - 5sinx + 7cosx - 2 with respect to x. Differentiating term by term: d/dx(3x⁴) = 12x³; d/dx(-5sinx) = -5cosx; d/dx(7cosx) = -7sinx; d/dx(-2) = 0. Adding these, f'(x) = 12x³ - 5cosx - 7sinx.