Trigonometric Limits and Theorems Notes - Limits Theorem

Continuity and Principles of Trigonometric Limits

  • The sine and cosine functions are characterized by continuity at every real number.

  • The tangent, cotangent, secant, and cosecant functions are continuous on their respective domains.

  • Verification of trigonometric limits using a calculator requires switching the device to radian mode. This is because the angle measure is based on arc length.

Fundamental Theorems and Corollaries

  • Theorem 1:     limx0sin(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1

  • Corollary to Theorem 1:     limx0xsin(x)=1\lim_{x \to 0} \frac{x}{\sin(x)} = 1

  • Theorem 2:     limx01cos(x)x=0\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0

  • Useful Trigonometric Identities:

    • Pythagorean Identity adaptation: 1cos2(x)=sin2(x)1 - \cos^2(x) = \sin^2(x)

    • Half-angle identity variation: 1cos(x)=2sin2(x2)1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right)

Sample Problem 1: Evaluating Limits with Sine

  • Problem: Find limx0sin(4x)x\lim_{x \to 0} \frac{\sin(4x)}{x}

  • Solution Procedure:

    • To utilize Theorem 1, the argument of the sine function must match the denominator. Multiply both the numerator and the denominator by 44.

    • The expression becomes: 4×limx0sin(4x)4x4 \times \lim_{x \to 0} \frac{\sin(4x)}{4x}

    • Because limu0sin(u)u=1\lim_{u \to 0} \frac{\sin(u)}{u} = 1, where u=4xu = 4x, the limit part evaluates to 11.

    • Calculation: 4×1=44 \times 1 = 4

    • The required solution is 44.

Sample Problem 2: Limit involving Reciprocal Sine

  • Problem: Find limx02xsin(3x)\lim_{x \to 0} \frac{2x}{\sin(3x)}

  • Solution Procedure:

    • Rewrite the expression to isolate the constant and apply the Corollary to Theorem 1: limx0xsin(x)=1\lim_{x \to 0} \frac{x}{\sin(x)} = 1.

    • Adjustment: 23×limx03xsin(3x)\frac{2}{3} \times \lim_{x \to 0} \frac{3x}{\sin(3x)}

    • By apply the limit property: 23×1=23\frac{2}{3} \times 1 = \frac{2}{3}

    • The required solution is 23\frac{2}{3}.

Sample Problem 3: Limit with Squared Sine and Constant Denominator

  • Problem: Find limx0sin2(2x)0.5x2\lim_{x \to 0} \frac{\sin^2(2x)}{0.5x^2}

  • Solution Procedure:

    • Rewrite the expression to handle the squared function: limx010.5×sin2(2x)x2\lim_{x \to 0} \frac{1}{0.5} \times \frac{\sin^2(2x)}{x^2}

    • Recognize that 10.5=2\frac{1}{0.5} = 2. Apply the Product Rule for limits: 2×(limx0sin(2x)x)22 \times \left(\lim_{x \to 0} \frac{\sin(2x)}{x}\right)^2

    • Evaluate the inner limit by multiplying by 22\frac{2}{2}: 2×(2×limx0sin(2x)2x)22 \times \left(2 \times \lim_{x \to 0} \frac{\sin(2x)}{2x}\right)^2

    • Calculation: 2×(2×1)2=2×4=82 \times (2 \times 1)^2 = 2 \times 4 = 8

    • The required solution is 88.

Sample Problem 4: Limit of Product with Tangent

  • Problem: Find limt0sin(t)tan(t)t\lim_{t \to 0} \frac{\sin(t) \tan(t)}{t}

  • Solution Procedure:

    • Factor the expression to separate known limits: limt0(sin(t)t×tan(t))\lim_{t \to 0} \left(\frac{\sin(t)}{t} \times \tan(t)\right)

    • Apply limit properties: (limt0sin(t)t)×(limt0tan(t))\left(\lim_{t \to 0} \frac{\sin(t)}{t}\right) \times \left(\lim_{t \to 0} \tan(t)\right)

    • Simplify using Theorem 1 and direct substitution: 1×tan(0)1 \times \tan(0)

    • Calculation: 1×0=01 \times 0 = 0

    • The required solution is 00.

Sample Problem 5: Applying Theorem 2

  • Problem: Find limx01cos(2x)4x\lim_{x \to 0} \frac{1 - \cos(2x)}{4x}

  • Solution Procedure:

    • Apply the form found in Theorem 2: limx01cos(u)u=0\lim_{x \to 0} \frac{1 - \cos(u)}{u} = 0.

    • Rewrite the expression: 12×limx01cos(2x)2x\frac{1}{2} \times \lim_{x \to 0} \frac{1 - \cos(2x)}{2x}

    • By applying the Product Rule: 12×0=0\frac{1}{2} \times 0 = 0

    • The required solution is 00.

Sample Problem 6: Using Identities for Squared Cosine Limits

  • Problem: Find limx01cos2(12x)3x2\lim_{x \to 0} \frac{1 - \cos^2(\frac{1}{2}x)}{3x^2}

  • Solution Procedure:

    • Use the identity 1cos2(x)=sin2(x)1 - \cos^2(x) = \sin^2(x) to rewrite the numerator as sin2(12x)\sin^2\left(\frac{1}{2}x\right).

    • Expression: limx0sin2(12x)3x2\lim_{x \to 0} \frac{\sin^2(\frac{1}{2}x)}{3x^2}

    • To apply the limit theorem, adjust the denominator to match the square of 12x\frac{1}{2}x. Since (12x)2=14x2\left(\frac{1}{2}x\right)^2 = \frac{1}{4}x^2, we transform the expression:

    • 13×limx0sin2(12x)x2\frac{1}{3} \times \lim_{x \to 0} \frac{\sin^2(\frac{1}{2}x)}{x^2}

    • 13×limx0(sin(x2)2×x2)2\frac{1}{3} \times \lim_{x \to 0} \left(\frac{\sin(\frac{x}{2})}{2 \times \frac{x}{2}}\right)^2

    • 13×(12)2×limx0(sin(x2)x2)2\frac{1}{3} \times \left(\frac{1}{2}\right)^2 \times \lim_{x \to 0} \left(\frac{\sin(\frac{x}{2})}{\frac{x}{2}}\right)^2

    • Solve and simplify: 13×14×(1)2=112\frac{1}{3} \times \frac{1}{4} \times (1)^2 = \frac{1}{12}

    • The required solution is 112\frac{1}{12}.

Sample Problem 7: Limit involving Tangent quotient

  • Problem: Find limx04xtan(x)\lim_{x \to 0} \frac{4x}{\tan(x)}

  • Solution Procedure:

    • Use the identity tan(x)=sin(x)cos(x)\tan(x) = \frac{\sin(x)}{\cos(x)} to rewrite the limit.

    • limx04xsin(x)cos(x)=limx0(4x×cos(x)sin(x))\lim_{x \to 0} \frac{4x}{\frac{\sin(x)}{\cos(x)}} = \lim_{x \to 0} \left(4x \times \frac{\cos(x)}{\sin(x)}\right)

    • Rearrange the terms: limx0[4cos(x)]×limx0[xsin(x)]\lim_{x \to 0} [4 \cos(x)] \times \lim_{x \to 0} [\frac{x}{\sin(x)}]

    • Since the cosine function is continuous, limx0cos(x)=cos(0)=1\lim_{x \to 0} \cos(x) = \cos(0) = 1.

    • Using the Corollary to Theorem 1: [4×cos(0)]×(1)=4×1×1=4[4 \times \cos(0)] \times (1) = 4 \times 1 \times 1 = 4

    • The required solution is 44.

Sample Problem 8: Combined Reciprocal Trigonometric Limits

  • Problem: Find limx0xcsc(3x)cot(x)\lim_{x \to 0} \frac{x \csc(3x)}{\cot(x)}

  • Solution Procedure:

    • Convert functions using identities: csc(3x)=1sin(3x)\csc(3x) = \frac{1}{\sin(3x)} and cot(x)=cos(x)sin(x)\cot(x) = \frac{\cos(x)}{\sin(x)}.

    • Expression: limx0x(1sin(3x))cos(x)sin(x)=limx0(xsin(3x)×sin(x)cos(x))\lim_{x \to 0} \frac{x \left(\frac{1}{\sin(3x)}\right)}{\frac{\cos(x)}{\sin(x)}} = \lim_{x \to 0} \left(\frac{x}{\sin(3x)} \times \frac{\sin(x)}{\cos(x)}\right)

    • Split the limit and adjust for Theorem 1 and its Corollary:

    • [13×limx03xsin(3x)]×[limx0sin(x)x×x]×[limx01cos(x)]\left[\frac{1}{3} \times \lim_{x \to 0} \frac{3x}{\sin(3x)}\right] \times \left[\lim_{x \to 0} \frac{\sin(x)}{x} \times x\right] \times \left[\lim_{x \to 0} \frac{1}{\cos(x)}\right]

    • Wait, properly simplifying the ratio: limx0[13×3xsin(3x)×sin(x)x×1cos(x)]\lim_{x \to 0} \left[\frac{1}{3} \times \frac{3x}{\sin(3x)} \times \frac{\sin(x)}{x} \times \frac{1}{\cos(x)}\right]

    • Solve and simplify: [13]×(1)×(1)×[1cos(0)]=13×1×1×1=13\left[\frac{1}{3}\right] \times (1) \times (1) \times \left[\frac{1}{\cos(0)}\right] = \frac{1}{3} \times 1 \times 1 \times 1 = \frac{1}{3}

    • The required solution is 13\frac{1}{3}.