Limits And Continuity — Flashcards
Telangana Open School (TOSS) · Class 12 · Mathematics
25 flashcards for Limits And Continuity (Telangana Open School (TOSS) Class 12 Mathematics) to test yourself on key terms and facts. Sample: "Evaluate: \"
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Evaluate: \(\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)
Answer
Step 1: Factor numerator: \(x^2 - 9 = (x - 3)(x + 3)\) Step 2: Simplify: \(\frac{(x - 3)(x + 3)}{x - 3} = x + 3\), for \(x \neq 3\) Step 3: Substitute: \(\lim_{x \to 3} (x + 3) = 3 + 3 = 6\) Answer…
Evaluate: \(\lim_{x \to 0} \frac{\sin 5x}{x}\)
Answer
Step 1: Use standard limit: \(\lim_{x \to 0} \frac{\sin kx}{x} = k\) Step 2: Rewrite: \(\frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x}\) Step 3: As \(x \to 0\), \(5x \to 0\), so \(\lim_{x \to 0} \fr…
Evaluate: \(\lim_{x \to 2} \frac{x^3 - 8}{x - 2}\)
Answer
Step 1: Recognize difference of cubes: \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\) Step 2: Cancel \((x - 2)\): \(\frac{(x - 2)(x^2 + 2x + 4)}{x - 2} = x^2 + 2x + 4\) Step 3: Substitute \(x = 2\): \(2^2 + 2(…
Evaluate: \(\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}\)
Answer
Step 1: Use identity: \(1 - \cos 2x = 2\sin^2 x\) Step 2: Rewrite: \(\frac{2\sin^2 x}{x^2} = 2 \left(\frac{\sin x}{x}\right)^2\) Step 3: Use standard limit: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) …
Evaluate: \(\lim_{x \to 0} \frac{e^{3x} - 1}{x}\)
Answer
Step 1: Use standard limit: \(\lim_{x \to 0} \frac{e^{kx} - 1}{x} = k\) Step 2: Rewrite: \(\frac{e^{3x} - 1}{x} = 3 \cdot \frac{e^{3x} - 1}{3x}\) Step 3: As \(x \to 0\), \(3x \to 0\), so \(\lim_{3x …
Evaluate: \(\lim_{x \to 1} \frac{x^4 - 1}{x - 1}\)
Answer
Step 1: Factor numerator: \(x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)\) Step 2: Cancel \((x - 1)\): \(\frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1} = (x + 1)(x^2 + 1)\) Step 3: Substitute \(x =…
Evaluate: \(\lim_{x \to 0} \frac{\tan x}{x}\)
Answer
Step 1: Write \(\tan x = \frac{\sin x}{\cos x}\) Step 2: \(\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}\) Step 3: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), \(\lim_{x \to 0} \frac{1}{\co…
Evaluate: \(\lim_{x \to \infty} \frac{3x^2 + 2x + 1}{5x^2 + 4}\)
Answer
Step 1: Divide numerator and denominator by \(x^2\): \(\frac{3 + \frac{2}{x} + \frac{1}{x^2}}{5 + \frac{4}{x^2}}\) Step 2: As \(x \to \infty\), \(\frac{2}{x} \to 0\), \(\frac{1}{x^2} \to 0\), \(\fra…
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