📝 Limit of 1 - cos x over x as x approaches 0 (15 MCQs)
📖 From Calculus • 2. Limits and Continuity an Introduction • 15 questions available
What is Limit of 1 - cos x over x as x approaches 0?
Definition:
The limit is another fundamental trigonometric limit, often derived using the identity and the known limit . This result shows that near zero, , making the numerator vanish faster than , which is essential for derivative calculations of cosine.
Example:
Evaluate .
Solution: Use identity: .
Reason:
This limit is essential for deriving the derivative of and for analyzing second-order effects in oscillations, making it a key tool in both pure mathematics and applied fields like signal processing.
📝 All Limit of 1 - cos x over x as x approaches 0 MCQs
Q1. What is \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos x}{x}\?
📖 Explanation: Using the series expansion \\\cos x=1-\\frac{x^{2}}{2}+\\dots\, the numerator behaves like \\\frac{x^{2}}{2}\. Dividing by \x\ gives \\\frac{x}{2}\, which tends to 0 as \x\\to0\. Hence the limit equals 0.
Q2. Which inequality can be used with the Squeeze Theorem to evaluate \\\lim_{x\\to 0}\\frac{1-\\cos x}{x}\?
📖 Explanation: For small \x\, the identity \1-\\cos x = 2\\sin^{2}(x/2)\ gives \0\\le 1-\\cos x\\le \\frac{x^{2}}{2}\. Dividing by \|x|\ yields \-\\frac{|x|}{2}\\le \\frac{1-\\cos x}{x}\\le \\frac{|x|}{2}\. The squeeze between \-|x|/2\ and \|x|/2\ forces the limit to 0.
Q3. Which Maclaurin series term is needed to find the limit of \\\frac{1-\\cos x}{x}\ as \x\\to0\?
📖 Explanation: The expression \1-\\cos x\ expands as \\\frac{x^{2}}{2}-\\frac{x^{4}}{24}+\\dots\. Keeping the first non‑zero term \\\frac{x^{2}}{2}\ is sufficient because after division by \x\ the dominant behavior is \\\frac{x}{2}\. Thus the cosine series term is essential for the limit.
Q4. Evaluate \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos x}{x^{2}}\.
📖 Explanation: Expanding \\\cos x\ gives \1-\\cos x = \\frac{x^{2}}{2}+O(x^{4})\. Dividing by \x^{2}\ leaves \\\frac{1}{2}+O(x^{2})\. As \x\\to0\ the higher‑order term vanishes, so the limit equals \\\frac{1}{2}\.
Q5. Let \f(x)=\\frac{1-\\cos x}{x}\ and \g(x)=\\sin x\. What is \\\displaystyle\\lim_{x\\to 0}\\frac{f(x)}{g(x)}\?
📖 Explanation: Using the series, \f(x)=\\frac{x}{2}+O(x^{3})\ and \g(x)=x+O(x^{3})\. Their quotient is \\\frac{\\frac{x}{2}+O(x^{3})}{x+O(x^{3})}=\\frac{1}{2}+O(x^{2})\. Hence as \x\\to0\ the limit approaches \\\frac{1}{2}\.
Q6. Which theorem directly justifies that \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos x}{x}=0\?
📖 Explanation: From the inequality \-\\frac{|x|}{2}\\le \\frac{1-\\cos x}{x}\\le \\frac{|x|}{2}\ we see the expression is trapped between two functions that both converge to 0. The Squeeze Theorem states that if a function is bounded above and below by two functions with the same limit, it shares that limit, giving the result 0.
Q7. Applying L'Hôpital's Rule, what is \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos x}{x}\?
📖 Explanation: Both numerator and denominator approach 0, allowing L'Hôpital's Rule. Differentiating gives \\\frac{\\sin x}{1}\. The limit of \\\sin x\ as \x\\to0\ is 0, so the original limit equals 0.
Q8. For a constant \k\, what is \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos(kx)}{x}\?
📖 Explanation: Using the series \1-\\cos(kx)=\\frac{k^{2}x^{2}}{2}+O(x^{4})\. Dividing by \x\ yields \\\frac{k^{2}x}{2}+O(x^{3})\, which tends to 0 as \x\\to0\. Thus the limit is 0 for any finite constant \k\.
Q9. Near \x=0\, the function \\\frac{1-\\cos x}{x}\ behaves like which of the following functions?
📖 Explanation: Expanding the numerator gives \1-\\cos x=\\frac{x^{2}}{2}+O(x^{4})\. Dividing by \x\ leaves \\\frac{x}{2}+O(x^{3})\, which is a linear function in \x\. Therefore the dominant behavior is linear, not constant or higher order.
Q10. Given \f(x)=\\frac{1-\\cos x}{x}\ for \x\\neq0\ and \f(0)=0\, which statement about continuity at \x=0\ is correct?
📖 Explanation: The original expression has a removable discontinuity at 0 because the limit exists (0) but the function is undefined there. By defining \f(0)=0\ we fill the hole, making the function continuous at \x=0\.
Q11. Let \a_n=\\frac{1-\\cos(1/n)}{1/n}\. What is \\\displaystyle\\lim_{n\\to\\infty}a_n\?
📖 Explanation: Replace \x=1/n\. As \n\\to\\infty\, \x\\to0\. The limit \\\lim_{x\\to0}\\frac{1-\\cos x}{x}\ is 0, so the sequence converges to 0.
Q12. Define \g(x)=\\int_{0}^{x}\\frac{1-\\cos t}{t}\\,dt\. As \x\\to0^{+}\, \g(x)\ approaches
📖 Explanation: For small \t\, \1-\\cos t\\approx t^{2}/2\, so the integrand behaves like \t/2\. Integrating \t/2\ from 0 to \x\ gives \x^{2}/4\, which tends to 0 as \x\\to0^{+}\. Hence the integral approaches 0.
Q13. Find \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos x}{\\sin x}\.
📖 Explanation: Using series, \1-\\cos x=\\frac{x^{2}}{2}+O(x^{4})\ and \\\sin x=x+O(x^{3})\. Their ratio is \\\frac{\\frac{x^{2}}{2}+O(x^{4})}{x+O(x^{3})}=\\frac{x}{2}+O(x^{3})\, which tends to 0 as \x\\to0\.
Q14. Compute \\\displaystyle\\lim_{x\\to 0}\\frac{1-\\cos x-\\frac{x^{2}}{2}}{x^{4}}\.
📖 Explanation: Subtracting the second‑order term gives \1-\\cos x-\\frac{x^{2}}{2}= -\\frac{x^{4}}{24}+O(x^{6})\. Dividing by \x^{4}\ yields \-\\frac{1}{24}+O(x^{2})\. Taking the limit as \x\\to0\ leaves \-\\frac{1}{24}\.
Q15. As \x\\to0\, \\\frac{1-\\cos x}{x}\ is asymptotically equivalent to
📖 Explanation: From the Maclaurin expansion, \1-\\cos x = \\frac{x^{2}}{2}+O(x^{4})\. Dividing by \x\ gives \\\frac{x}{2}+O(x^{3})\. The leading term \\\frac{x}{2}\ captures the dominant behavior, making it asymptotically equivalent to \\\frac{x}{2}\.