Limits

Answer

For a function f(x) and constants a and l, we say lim(x→a) f(x) = l if: Given any ε > 0, there exists δ > 0 such that |f(x) - l| < ε whenever 0 < |x - a| < δ. This means f(x) can be made arbitrarily c

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Step 1: Direct substitution gives 0/0 (indeterminate form) Step 2: Factor the numerator: x² - 9 = (x + 3)(x - 3) Step 3: Simplify: (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3 (for x ≠ 3) Step 4:

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If lim(x→a) f(x) = l and lim(x→a) g(x) = m, then: 1. lim(x→a) [f(x) ± g(x)] = l ± m 2. lim(x→a) [f(x) × g(x)] = l × m 3. lim(x→a) [k·f(x)] = k·l (k constant) 4. lim(x→a) [f(x)/g(x)] = l/m (if m ≠ 0) 5

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Step 1: Direct substitution gives 0/0 form Step 2: Multiply by conjugate: (√(1+x) - 1)/x × (√(1+x) + 1)/(√(1+x) + 1) Step 3: Simplify numerator: (1+x) - 1 = x Step 4: Expression becomes: x/[x(√(1+x) +

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Theorem: lim(θ→0) (sin θ)/θ = 1 (θ in radians) Proof using Squeeze Theorem: For 0 < θ < π/2, consider a unit circle: Area of triangle OAP < Area of sector OAP < Area of triangle OAB ½·sin θ < ½·θ < ½

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Step 1: Rewrite using basic trigonometric identities sin 5x/(tan 3x) = (sin 5x)/(sin 3x/cos 3x) = (sin 5x × cos 3x)/(sin 3x) Step 2: Multiply and divide by x = (sin 5x/x) × (x/sin 3x) × cos 3x × (5x/5

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Standard Limit: lim(x→0) (e^x - 1)/x = 1 Related Forms: 1. lim(x→0) (a^x - 1)/x = log a (a > 0, a ≠ 1) 2. lim(x→0) (1 + x)^(1/x) = e 3. lim(x→0) log(1 + x)/x = 1 Significance: These limits are funda

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Step 1: Split the fraction (5^x - 3^x)/x = (5^x - 1)/x - (3^x - 1)/x Step 2: Apply standard limit formula lim(x→0) (a^x - 1)/x = log a Step 3: Evaluate each term lim(x→0) (5^x - 1)/x = log 5 lim(x→0)