Let an= ∫ limits-ππ |x-1| cos nx dx for all natural numbers n Then sequence (an)n ge 0 satisfies

Question

Let an= ∫ limits-ππ |x-1| cos nx dx for all natural numbers n Then sequence (an)n ge 0 satisfies (A)  displaystyle limn→∞ an =∞ (B)  displaystyle limn→∞ an =-∞ (C)  displaystyle limn→∞ an exists and is positive (D)  displaystyle limn→∞ an =0

Tardigrade Admin · Accepted Answer

We have, an=∫ limits-ππ|x-1| cos n x d x an=∫ limits-π1-(x-1) cos n x d x -∫ limits1π(x-1) cos n x d x an=-[((x-1) sin n x/n)]-π1-[( cos n π/n2)]-π1 +[((x-1) sin n x/n)]1π-[( cos n π/n2)]1π an=(2 π sin n π/n)+(2/n2) cos n π-(2/n2) cos x π an=(2 π sin n π/n) displaystyle lim x arrow ∞ an= displaystyle lim n arrow ∞ (2 π sin n π/n)=0

Q. Let $a_{n} = - π \int π ∣ x - 1 ∣ cos n x d x$ for all natural numbers $n$ Then sequence $(a_{n})_{n \geq 0}$ satisfies

$n \to \infty lim a_{n} = \infty$

$n \to \infty lim a_{n} = - \infty$

$n \to \infty lim a_{n}$ exists and is positive

$n \to \infty lim a_{n} = 0$

Q. Let an​=−π∫π​∣x−1∣cosnxdx for all natural numbers n Then sequence (an​)n≥0​ satisfies

n→∞lim​an​=∞

n→∞lim​an​=−∞

n→∞lim​an​ exists and is positive

n→∞lim​an​=0

Q. Let $a_{n} = - π \int π ∣ x - 1 ∣ cos n x d x$ for all natural numbers $n$ Then sequence $(a_{n})_{n \geq 0}$ satisfies

$n \to \infty lim a_{n} = \infty$

$n \to \infty lim a_{n} = - \infty$

$n \to \infty lim a_{n}$ exists and is positive

$n \to \infty lim a_{n} = 0$