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).

  • $\begingroup$ Can you please clarify how did you arrive at the last step from the second last? $\endgroup$
    – user1012971
    Jan 26, 2022 at 17:13
  • $\begingroup$ @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... $\endgroup$
    – DeBARtha
    Jan 26, 2022 at 17:48

2 Answers 2


Your method is incorrect since

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

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


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

Therefore the limit is indeed $2$

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

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jan 26, 2022 at 17:27

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