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Calculus - Limits of Polynomials, Rational, Trigonometric, Exponential and Logarithmic Functions

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of a Limit: The limit of a function f(x)f(x) as xx approaches aa is the value LL that f(x)f(x) gets closer to from both the left side (x→a−x \to a^{-}) and the right side (x→a+x \to a^{+}). Visually, if you trace the graph from both sides, the yy-coordinates must converge toward the same height LL for the limit to exist, even if there is a 'hole' in the graph at x=ax = a.

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Direct Substitution for Polynomials: For any polynomial function P(x)P(x), the limit as x→ax \to a is simply P(a)P(a). Visually, polynomial graphs are smooth, continuous curves without any breaks or jumps, meaning the value the function 'should' reach is exactly the value it 'does' reach.

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Rational Functions and Indeterminate Forms: For a function f(x)g(x)\frac{f(x)}{g(x)}, if direct substitution results in 00\frac{0}{0}, it is called an indeterminate form. Visually, this usually signifies a removable discontinuity (a hole) in the graph. You must simplify the expression by factoring and canceling common terms before substituting the value again.

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Limits of Trigonometric Functions: These involve specialized identities like lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Visually, as xx (measured in radians) gets closer to zero, the length of the vertical segment representing sin⁡x\sin x on a unit circle becomes almost identical to the arc length xx, causing their ratio to approach 1.

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Exponential and Logarithmic Growth: These limits describe the behavior of growth functions near zero. For example, lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1. Visually, this indicates that the slope of the curve y=exy = e^x at the point (0,1)(0, 1) is exactly 1, meaning the curve is tangent to the line y=x+1y = x + 1 at that specific point.

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Existence of a Limit: A limit exists at x=ax = a if and only if the Left-Hand Limit (LHL) equals the Right-Hand Limit (RHL), i.e., lim⁡x→a−f(x)=lim⁡x→a+f(x)\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x). If the graph shows a 'jump' (like a step function) where the left side ends at one height and the right side starts at another, the limit does not exist.

📐Formulae

lim⁡x→a[f(x)±g(x)]=lim⁡x→af(x)±lim⁡x→ag(x)\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)

lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}

lim⁡x→0sin⁡xx=1 (where x is in radians)\lim_{x \to 0} \frac{\sin x}{x} = 1 \text{ (where } x \text{ is in radians)}

lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1

lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0

lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1

lim⁡x→0ax−1x=log⁡ea or ln⁡a\lim_{x \to 0} \frac{a^x - 1}{x} = \log_e a \text{ or } \ln a

lim⁡x→0log⁡e(1+x)x=1\lim_{x \to 0} \frac{\log_e(1+x)}{x} = 1

💡Examples

Problem 1:

Evaluate the limit: lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Solution:

  1. Direct substitution gives 32−93−3=00\frac{3^2 - 9}{3 - 3} = \frac{0}{0}, which is indeterminate.
  2. Factor the numerator: x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3).
  3. Rewrite the limit: lim⁡x→3(x−3)(x+3)x−3\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}.
  4. Cancel the common factor (x−3)(x - 3): lim⁡x→3(x+3)\lim_{x \to 3} (x + 3).
  5. Substitute x=3x = 3: 3+3=63 + 3 = 6.

Explanation:

This is a rational function limit. Since substitution resulted in 0/00/0, we used the factorization method to remove the 'hole' at x=3x=3 and find the value the function was approaching.

Problem 2:

Evaluate the limit: lim⁡x→0sin⁡4x3x\lim_{x \to 0} \frac{\sin 4x}{3x}

Solution:

  1. We know the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1.
  2. To make the argument of sine match the denominator, multiply and divide the expression by 4: 44⋅sin⁡4x3x\frac{4}{4} \cdot \frac{\sin 4x}{3x}.
  3. Rearrange the terms: 43⋅sin⁡4x4x\frac{4}{3} \cdot \frac{\sin 4x}{4x}.
  4. Apply the limit: 43⋅lim⁡x→0sin⁡4x4x\frac{4}{3} \cdot \lim_{x \to 0} \frac{\sin 4x}{4x}.
  5. Since 4x→04x \to 0 as x→0x \to 0, the limit becomes 43⋅1=43\frac{4}{3} \cdot 1 = \frac{4}{3}.

Explanation:

This trigonometric limit is solved by manipulating the expression to match the standard identity sin⁡θθ\frac{\sin \theta}{\theta}. We adjusted the denominator to match the angle 4x4x and extracted the constant coefficient.