If x1-3 and xn-1-2-xn- n- 1- then n-lim xn is-

X_n+1=√(2+x_n) ⇒n→∞lim x_n+1=√(2+n→∞lim x_n) nbsp; t=√(2+t) (∵n→∞lim x_n+1=n→∞lim x_n=t ) ⇒ t^2 t 2=0 nbsp; ⇒(t 2)(t+1)=0⇒ nbsp; t=2 nbsp; (∵ t gt;0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Summation Using Sigma

If x1=3 and x...

Question

If $x_{1} = 3$ and $x_{n + 1} = \sqrt{2 + x_{n}}, n \geq 1$ , then $lim n \to \infty x_{n}$ is

Open in App

Solution

$x_{n + 1} = \sqrt{2 + x_{n}}$
$\Rightarrow lim n \to \infty x_{n + 1} = \sqrt{2 + lim n \to \infty x_{n}}$
$t = \sqrt{2 + t} (∵ lim n \to \infty x_{n + 1} = lim n \to \infty x_{n} = t)$
$\Rightarrow t^{2} - t - 2 = 0$
$\Rightarrow (t - 2) (t + 1) = 0 \Rightarrow t = 2 (∵ t > 0)$

Suggest Corrections

Similar questions

Q. If

x_{1} = 3

and

x_{n + 1} = \sqrt{2 + x_{n}}, n \geq 1

, then

lim n \to \infty x_{n}

Q. If

x

is a positive number. A sequence is defined as

x_{1} = \frac{1}{2} (x + \frac{3}{x}), x_{2} = \frac{1}{2} (x_{1} + \frac{3}{x_{1}}) . . . x_{n + 1} = \frac{1}{2} (x_{n} + \frac{3}{x_{n}}), n \geq 1,

then

lim n \to \infty x_{n}

is equal to

$The value of l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1}$

Q. If

x_{1}, x_{2}, x_{3}, x_{4} . . . . x_{2 n + 1}

are in Arithmetic Progression, then find the value of

[\frac{(x_{2 n + 1} - x_{1})}{(x_{2 n + 1} + x_{1})}] + [\frac{(x_{2 n} - x_{2})}{(x_{2 n} + x_{2})}] . . . . . . . . . . . . . . . . . . . . [\frac{(x_{n + 2 - x_{n}})}{(x_{n + 2} + x_{n})}]

Q. Let

x_{1} = 1, x_{3} = 4

and

x_{n + 3} = 2 x_{n + 2} + 2 x_{n + 2} - x_{n}

for all

n \geq 1.

Prove that

x_{n}

is a square for all

n \geq 1.

Join BYJU'S Learning Program

Explore more

Summation Using Sigma

Standard XII Mathematics

Join BYJU'S Learning Program

If x1=3 and xn+1=√2+xn,n≥1, then limn→∞xn is​​​​​​​

xn+1=√2+xn ⇒limn→∞xn+1=√2+limn→∞xn t=√2+t (∵limn→∞xn+1=limn→∞xn=t) ⇒t2−t−2=0 ⇒(t−2)(t+1)=0⇒t=2(∵t>0)

If $x_{1} = 3$ and $x_{n + 1} = \sqrt{2 + x_{n}}, n \geq 1$ , then $lim n \to \infty x_{n}$ is

$x_{n + 1} = \sqrt{2 + x_{n}}$
$\Rightarrow lim n \to \infty x_{n + 1} = \sqrt{2 + lim n \to \infty x_{n}}$
$t = \sqrt{2 + t} (∵ lim n \to \infty x_{n + 1} = lim n \to \infty x_{n} = t)$
$\Rightarrow t^{2} - t - 2 = 0$
$\Rightarrow (t - 2) (t + 1) = 0 \Rightarrow t = 2 (∵ t > 0)$