# Calculating $\lim_{n\to\infty}\left(2^n+n2^n\sin^2{\frac{n}{2}}\right)^{\left(\frac{1}{2n-n\cos{\frac{1}{n}}}\right)}$

I was looking up to some previous papers of a competitive exam when I found the following problem: $$\lim_{n\to\infty}\left(2^n+n\cdot2^n\sin^2{\frac{n}{2}}\right)^{\left(\frac{1}{2n-n\cos{\frac{1}{n}}}\right)}$$

I proceeded as follows: $$\require{cancel}\lim_{n\to\infty}\left(2^n+n\cdot2^n\sin^2{\frac{n}{2}}\right)^{\left(\frac{1}{2n-n\cos{\frac{1}{n}}}\right)}\\ =\lim_{n\to\infty}2^\left(\frac{1}{2-\cos{\frac{1}{n}}}\right)\cdot\lim_{n\to\infty}\left(1+n\cdot\sin^2{\frac{n}{2}}\right)^{\left(\frac{1}{2n-n\cos{\frac{1}{n}}}\right)}\\ =\cancelto{2^1=2}{\lim_{n\to\infty}2^\left(\frac{1}{2-\cos{\frac{1}{n}}}\right)}\cdot\lim_{n\to\infty}\left(1+n\cdot\sin^2{\frac{n}{2}}\right)^{\left(\frac{1}{2n-n\cos{\frac{1}{n}}}\right)}\\ =2\cdot\lim_{n\to\infty}\left(1+n\cdot\sin^2{\frac{n}{2}}\right)^{\left(\frac{1}{2n-n\cos{\frac{1}{n}}}\right)}$$

After this step, my intuition was that the answer is $$2$$ as this was an integer-only answer type question. But, I cannot seem to proceed after this step. I did apply the L'hospital method after this, but, I didn't go far. A little bit of hint/help would be very kind.

P.S.: WolframAlpha did confirm my intuition (see here).

• Can you please clarify how did you arrive at the last step from the second last?
– user1012971
Jan 26, 2022 at 17:13
• @RamanujanXV I tried to use the $1^\infty$ form. However, I made a mistake here, as DatBoi mentioned in his answer below. I'll edit and cut out that part from the question... Jan 26, 2022 at 17:48

$$\lim_{n\to\infty}1+n\cdot\sin^2{\frac{n}{2}}\ne 1$$

So you cannot possibly use the $$1^{\infty}$$ form

To solve the limit we can use logarithms

$$L=\lim_{n\to\infty}\left(1+n\cdot\sin^2{\frac{n}{2}}\right)^{\left(\frac{1} {2n-n\cos{\frac{1}{n}}}\right)}$$

$$\ln L=\lim_{n\to\infty}\frac{\ln\left(1+n\cdot\sin^2{\frac{n}{2}}\right)}{{2n-n\cos{\frac{1}{n}}}}=\lim_{n\to\infty}\frac{\ln\left(1+n\cdot\sin^2{\frac{n}{2}}\right)}{n}$$

Since $$\sin^2x\in [0,1]$$

$$\lim_{n\to\infty}\frac{\ln\left(1\right)}{n}\le\lim_{n\to\infty}\frac{\ln\left(1+n\cdot\sin^2{\frac{n}{2}}\right)}{n}\le\lim_{n\to\infty}\frac{\ln\left(1+n\right)}{n}$$

$$0\le\ln L\le0$$

Therefore the limit is indeed $$2$$

• I didn't quite get the fact how you managed to derive $n$ in the denominator. Can you clarify that once pls? Jan 26, 2022 at 17:51
• @DeBARtha $\lim_{n\to\infty}\cos(\frac{1}{n})=?$ Jan 26, 2022 at 17:55
• But there's a $2n$ in the denominator also, and $\cos\frac{1}{n}$ has $n$ multiplied to it. Won't that change also? Jan 26, 2022 at 17:58
• @DeBARtha $\lim_{n\to\infty}\cos(\frac{1}{n})$ takes a specific value that doesnt affect the $n$ thats multiplied with it Jan 26, 2022 at 18:00

look on the parts.

(2-cos1/n)n -> (2-cos0) n= (2-1)n = n

so the power is just 1/n

(2^n) ^1/n =2

1 <= (1+n*sin^2(n/2))^1/n <= (1+n)^1/n=1 ( * )

( * ) in detail is when you write (*) = e^(ln(1+n)/n) then apply Lopetals rule, bottom wins and you ver e^0 = 1

so the answer is 2*1 = 2