Question: Evaluate the following limit $$\lim _{n \rightarrow \infty}\left(\frac{\sin \left\{\frac{2}{n}\right\}}{\left[2 n \tan \frac{1}{n}\right]\left(\tan \frac{1}{n}\right)}+\frac{1}{n^{2}+\cos n}\right)^{n^{2}}$$ here {} and [] denote the fractional part function and the greatest integer function respectively.

Answer: The answer of this question is given as $1$, the problem is from an JEE Advanced practice problems set.

My approach: I figured out that this is the $1^{\infty}$ form, so I tried to convert it in the form $$e^{\lim_{{n}\rightarrow{\infty}}n^{2}.G(n)}$$ here $G(n)$ is the function within the brackets, after this step I am not able to proceed as the limit in the power of $e$ is very messy and not convertible into some standard form, please help.

  • $\begingroup$ You could perhaps use $\sin(2x) = \frac{2\tan(x)}{1 + \tan^2(x)}$ $\endgroup$ – Gribouillis Aug 22 '20 at 16:21
  • $\begingroup$ How do I deal with the fractional part function then? $\endgroup$ – Shriom707 Aug 22 '20 at 17:24

Using $\{x\}=x-\lfloor x\rfloor$, we have for $n>2$

$$\begin{align} \sin(\{2/n\})&=\sin\left(2/n-\lfloor2/n\rfloor\right)\\\\ &=\sin(2/n)\cos(\lfloor2/n\rfloor)-\cos(2/n)\sin(\lfloor2/n\rfloor)\\\\ &=\sin(2/n)\\\\ &=2\sin(1/n)\cos(1/n) \end{align}$$

In addition, for $n>2$, $\lfloor2n \tan(\frac1n)\rfloor=2$.

Hence, we can write for $n>2$

$$\begin{align} \left(\frac{\sin(\{2/n\})}{\lfloor2n \tan(\frac1n)\rfloor \tan(1/n)}+\frac1{n^2+\cos(n)}\right)^{n^2}&=\left(\cos^2(1/n)+\frac1{n^2+\cos(n)}\right)^{n^2}\\\\ &=\left(1+O\left(\frac1{n^4}\right)\right)^{n^2} \end{align}$$

whereupon letting $n\to \infty$ yields the coveted limit

$$\lim_{n\to\infty}\left(\frac{\sin(\{2/n\})}{\lfloor2n \tan(\frac1n)\rfloor \tan(1/n)}+\frac1{n^2+\cos(n)}\right)^{n^2}=1$$

  • $\begingroup$ a very elegant solution, thanks a lot $\endgroup$ – Shriom707 Aug 22 '20 at 17:23
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Aug 22 '20 at 17:24
  • $\begingroup$ how did you obtain the $n^{4}$ term in the third step of your solution? $\endgroup$ – Shriom707 Aug 23 '20 at 7:40
  • $\begingroup$ Note that $$\cos^2(1/n)=\left(1-\frac1{n^2}+O\left(\frac1{n^4}\right)\right)$$and$$ \frac1{n^2+\cos(n)}=\frac{1}{n^2}\left(1-\frac{\cos(n)}{n^2}+O\left(\frac1{n^4}\right)\right)=\frac1{n^2}+O\left(\frac1{n^4}\right)$$ $\endgroup$ – Mark Viola Aug 23 '20 at 15:38

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