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Existence of Limit

If limn→∞1/...

Question

If $lim n \to \infty \frac{1}{2^{n}} = L$

Then find the value of $L$ .

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Solution

As $n \to \infty$
$a^{n} \to 0$ where 0<a<1
$∴ lim x \to \infty \frac{1}{2^{n}} = {(\frac{1}{2})}^{n} = 0$

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Similar questions

L = lim n \to \infty n^{- n^{2}} {[(n + 1) (n + \frac{1}{2}) (n + \frac{1}{2^{2}}) \dots (n + \frac{1}{2^{n - 1}})]}^{n}

, then the value of

ln L

L = lim n \to \infty (2 \times 3^{2} \times 2^{3} \times 3^{4} . . . \times 2^{n - 1} \times 3^{n})^{\frac{1}{(n^{2} + 1)}}

, then the value of

L^{4}

a = lim n \to \infty \frac{tan (\frac{π}{2^{2}} + \frac{π}{2^{3}} + . . . . + \frac{π}{2^{n + 1}})}{2^{n - 1}}

, then value of

π a

l = lim n \to \infty n \sum r = 2 ((r + 1) sin \frac{π}{r + 1} - r sin \frac{π}{r})

{l}

{}

denotes the fractional part function.

A sequence is a function whose domain is the set of natural number. A sequence

s_{1}, s_{2} . . .

of real numbers is said to have a limit

l

{lim}_{n \to \infty} s_{n} = l

l < \infty,

< s_{n} >

is said to convergent. The following are well known
(i)

lim n \to \infty \frac{1}{n^{p}} = 0 (p > 0)

lim n \to \infty x^{n} = 0 (| x | < 1)

lim n \to \infty s_{n} = l

lim n \to \infty \frac{s_{1} + s_{2} + . . . + s_{n}}{n} = l

f (x) = - 1 + | x - 1 |, g (x) = 2 - | x + 1 |

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