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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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The limit of a function f(x)f(x) as xx approaches aa represents the value that f(x)f(x) gets closer to as xx moves toward aa from both the left and right sides. If the left-hand limit lim⁡x→a−f(x)\lim_{x \to a^-} f(x) and the right-hand limit lim⁡x→a+f(x)\lim_{x \to a^+} f(x) are equal, the limit exists.

Graph showing a function approaching a specific value L as x approaches a.
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For polynomial functions P(x)=anxn+...+a0P(x) = a_n x^n + ... + a_0, the limit is simply found by direct substitution: lim⁡x→aP(x)=P(a)\lim_{x \to a} P(x) = P(a). Rational functions P(x)Q(x)\frac{P(x)}{Q(x)} are also solved by substitution provided Q(a)≠0Q(a) \neq 0.

Graph of y = x^2 showing that the limit as x approaches 2 is 4.
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Trigonometric limits often involve the sandwich theorem or fundamental identities like lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. When xx is small and in radians, sin⁡x≈x\sin x \approx x.

Graph of sin(x)/x showing the hole at x=0 where the limit is 1.
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Exponential and logarithmic limits deal with the growth rates of functions. Key forms include lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1 and lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1.

Graph of the exponential function e^x.

📐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.

Problem 3:

Evaluate the limit: lim⁡x→2x3−8x2−4\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}

Graph of the simplified rational function approaching the value 3 at x=2.

Solution:

lim⁡x→2(x−2)(x2+2x+4)(x−2)(x+2)\lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)} Canceling the common factor (x−2):\text{Canceling the common factor } (x-2): lim⁡x→2x2+2x+4x+2\lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} By substitution: 22+2(2)+42+2=124=3\text{By substitution: } \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{12}{4} = 3

Explanation:

This rational function takes the form 0/00/0 at x=2x=2. We factorize the numerator using the difference of cubes formula a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2) and the denominator using the difference of squares formula.

Problem 4:

Evaluate the limit: lim⁡x→0e5x−1sin⁡2x\lim_{x \to 0} \frac{e^{5x} - 1}{\sin 2x}

Graph showing the function (e^{5x}-1)/sin(2x) approaching 2.5 as x approaches 0.

Solution:

Divide numerator and denominator by x:\text{Divide numerator and denominator by } x: lim⁡x→0e5x−1xsin⁡2xx\lim_{x \to 0} \frac{\frac{e^{5x} - 1}{x}}{\frac{\sin 2x}{x}} Multiply by constants to match standard forms:\text{Multiply by constants to match standard forms:} lim⁡x→05×e5x−15xlim⁡x→02×sin⁡2x2x\frac{\lim_{x \to 0} 5 \times \frac{e^{5x} - 1}{5x}}{\lim_{x \to 0} 2 \times \frac{\sin 2x}{2x}} 5×12×1=52=2.5\frac{5 \times 1}{2 \times 1} = \frac{5}{2} = 2.5

Explanation:

We use the standard limits lim⁡u→0eu−1u=1\lim_{u \to 0} \frac{e^u - 1}{u} = 1 and lim⁡v→0sin⁡vv=1\lim_{v \to 0} \frac{\sin v}{v} = 1 by adjusting the variables to match the coefficients of xx.