A sequence is defined by a_1 = 3 and a_{n+1} = sqrt(6 + a_n). Assuming the sequence converges, find its limit L.
Answer and explanation
Correct answer: C. 3
- Step 1: Take the limit on both sides: L = sqrt(6 + L).
- Step 2: Square both sides: L^2 = 6 + L, which rearranges to L^2 - L - 6 = 0.
- Step 3: Factor: (L - 3)(L + 2) = 0.
- Step 4: Since terms are positive square roots, L must be positive, hence L = 3.
Answer
3.
Why not the others: sqrt(6) is the first term after 0; 6 fails L^2 - L - 6 = 36 - 6 - 6 != 0; 2 gives 2^2 - 2 - 6 = -4 != 0.
Practise more Differential and Integral Calculus and Complex Numbers.
Free questions with an explanation for every one, and full timed Bagrut (Secondary School Matriculation) mock tests.
More Differential and Integral Calculus and Complex Numbers questions
- Solve the trigonometric equation 2sin(x)^2 - 3sin(x) + 1 = 0 on the interval [0, 2*pi].
- Which expression is identical to cos(2*x) / (cos(x) + sin(x)) where defined?
- Find all real solutions to sin(3*x) + sin(x) = cos(x) on the interval [0, pi].
- If sin(x) = 3/5 and x is in the second quadrant (pi/2 < x < pi), what is the exact value of tan(2*x)?
- What is the fundamental period of the function f(x) = 3tan(2x - pi/4) + 1?
- Determine all intersection points of y = 2cos(x) and y = 1/cos(x) on the interval [0, 2pi].