# Evaluate the following trigonometric limit:

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.

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

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

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