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Limits and Derivatives - Limits of Trigonometric Functions

Grade 11CBSE

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

🔑Concepts

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Direct Substitution: For basic trigonometric functions like sin⁡x\sin x and cos⁡x\cos x, the limit as x→ax \to a is simply the function's value at that point, provided it is defined. Visually, since sine and cosine are continuous wave functions, as you trace the curve towards a point, the height of the curve approaches the yy-value of that point.

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The Squeeze Theorem: This theorem is fundamental for proving trigonometric limits. If a function f(x)f(x) is 'sandwiched' between two other functions g(x)g(x) and h(x)h(x) that both approach the same limit LL at a point, then f(x)f(x) must also approach LL. On a graph, imagine two outer curves pinching an inner curve at a single point.

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The Fundamental Limit of Sine: The most important limit is lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Although the function is undefined at x=0x = 0 (creating a 'hole' at (0,1)(0, 1) on a graph), the ratio of the opposite side to the arc length approaches unity as the angle xx (in radians) gets infinitely small.

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Geometric Interpretation of sin⁡xx\frac{\sin x}{x}: On a unit circle, for a small angle xx, the vertical height (sin⁡x)(\sin x), the arc length (x)(x), and the tangent segment (tan⁡x)(\tan x) are very close in value. As xx shrinks to 00, the area of the triangle inside the sector and the sector itself become nearly identical, explaining why their ratio approaches 11.

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Indeterminate Forms: Many trigonometric limits initially result in 00\frac{0}{0} when using direct substitution. This indicates that the expression needs to be simplified using trigonometric identities or standard limit theorems to find the actual value.

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Trigonometric Transformations: To evaluate complex limits, we often use identities like 1−cos⁡x=2sin⁡2(x2)1 - \cos x = 2\sin^2(\frac{x}{2}). Visually, this transforms a function that behaves like a flat curve near the origin (1−cos⁡x)(1 - \cos x) into a squared sine function, which is easier to relate to the standard sin⁡xx\frac{\sin x}{x} limit.

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Substitution Method: When a limit variable xx approaches a value other than 00 (e.g., x→πx \to \pi), we use substitution. By letting y=x−πy = x - \pi, we can transform the problem into a new limit where y→0y \to 0, allowing us to use standard trigonometric limit formulas.

📐Formulae

lim⁡x→0sin⁡x=0\lim_{x \to 0} \sin x = 0

lim⁡x→0cos⁡x=1\lim_{x \to 0} \cos x = 1

lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 (where xx 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→01−cos⁡xx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

💡Examples

Problem 1:

Evaluate lim⁡x→0sin⁡5x3x\lim_{x \to 0} \frac{\sin 5x}{3x}

Solution:

  1. We know the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1.
  2. To use this, the angle in the sine function must match the denominator. Here the angle is 5x5x, but the denominator is 3x3x.
  3. Multiply and divide the expression by 55: lim⁡x→0sin⁡5x3x=lim⁡x→0(sin⁡5x5x⋅53)\lim_{x \to 0} \frac{\sin 5x}{3x} = \lim_{x \to 0} \left( \frac{\sin 5x}{5x} \cdot \frac{5}{3} \right)
  4. Pull the constant out of the limit: 53⋅lim⁡x→0sin⁡5x5x\frac{5}{3} \cdot \lim_{x \to 0} \frac{\sin 5x}{5x}
  5. Let y=5xy = 5x. As x→0x \to 0, yy also approaches 00. The limit becomes: 53⋅lim⁡y→0sin⁡yy=53⋅1=53\frac{5}{3} \cdot \lim_{y \to 0} \frac{\sin y}{y} = \frac{5}{3} \cdot 1 = \frac{5}{3}

Explanation:

This solution uses the strategy of coefficient adjustment to match the argument of the sine function with its denominator, allowing the application of the fundamental limit theorem.

Problem 2:

Evaluate lim⁡x→01−cos⁡4xx2\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}

Solution:

  1. Use the identity 1−cos⁡θ=2sin⁡2(θ2)1 - \cos \theta = 2\sin^2(\frac{\theta}{2}). Here θ=4x\theta = 4x, so θ2=2x\frac{\theta}{2} = 2x.
  2. Substitute the identity into the limit: lim⁡x→02sin⁡2(2x)x2\lim_{x \to 0} \frac{2\sin^2(2x)}{x^2}
  3. Rearrange the expression to group the squared terms: 2⋅lim⁡x→0(sin⁡2xx)22 \cdot \lim_{x \to 0} \left( \frac{\sin 2x}{x} \right)^2
  4. To use lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1, we need 2x2x in the denominator. Multiply and divide inside the square by 22: 2⋅lim⁡x→0(2⋅sin⁡2x2x)2=2⋅(2)2⋅lim⁡x→0(sin⁡2x2x)22 \cdot \lim_{x \to 0} \left( \frac{2 \cdot \sin 2x}{2x} \right)^2 = 2 \cdot (2)^2 \cdot \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \right)^2
  5. Evaluate the limit: 2⋅4⋅(1)2=82 \cdot 4 \cdot (1)^2 = 8

Explanation:

This approach uses a trigonometric identity to convert a cosine expression into a sine expression, which then permits the use of the standard sine limit by squaring the terms.