N- 04 limsin - - 2n - 2sin - - n - sin-n2-

4lim_n → 0sin( α #160; + 2n) 2sin( α #160; + n)+sinα/n^2 [ 0/0for n] (By L.H rule) 4lim_n → 02cos( α #160; + 2n) 2 cos( α #160; + n ...

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

Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

n→ 04 limsin ...

Question

$4 l i m n \to 0 \frac{sin (α + 2 n) - 2 sin (α + n) + sin α}{n^{2}}$ .

Open in App

Solution

$4 lim n \to 0 \frac{sin (α + 2 n) - 2 sin (α + n) + s i n α}{n^{2}} [\frac{0}{0} f o r n]$
(By L.H rule)
$4 lim n \to 0 \frac{2 cos (α + 2 n) - 2 cos (α + n) + 0}{2 n}$
$8 lim n \to 0 \frac{cos (α + 2 n) - cos (α + n)}{2 n}$
By. L. H. rule.
$8 lim n \to 0 \frac{- sin (α + 2 n) \times 2 + sin (α + n)}{2}$
$= 8 ∣ ∣ \frac{- sin (α + 0) + sin (α + 0)}{2} ∣ ∣$
Hence, We get,
$= 8 ∣ ∣ \frac{- sin α + s i n α}{2} ∣ ∣$
=0

Suggest Corrections

Similar questions

If \frac{^{n + 2} C_{6}}{^{n - 2} P_{2}} = 11

, then

n

satisfies the equation:

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

is equal to

n \sum r = 0 r^{2} . (^{n} C_{r})^{2}

is equal to

2ⁿ > n² when n Є N such that

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

Join BYJU'S Learning Program

4 limn→0sin(α+2n)−2sin(α+n)+sinαn2.

4limn→0sin(α+2n)−2sin(α+n)+sinαn2[00forn](By L.H rule)4limn→02cos(α+2n)−2cos(α+n)+02n8limn→0cos(α+2n)−cos(α+n)2nBy. L. H. rule.8limn→0−sin(α+2n)×2+sin(α+n)2=8∣∣−sin(α+0)+sin(α+0)2∣∣Hence, We get,=8∣∣−sinα+sinα2∣∣=0

$4 l i m n \to 0 \frac{sin (α + 2 n) - 2 sin (α + n) + sin α}{n^{2}}$ .