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Calculate limit $$\lim_{n\to\infty} \left(n \cdot \tan \left( \frac{3}{2n+2} \right)\right)$$ every suggestion by every way will be grateful.

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    The limit is $1$, simply by using the approximation $\tan x \sim x$ around $0$. – 2017-02-02
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    Are you asking what my contribution to solving the problem was? I am unable to calculate the limit mentioned.. – 2017-02-02
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    Are $t,g$ constants (or $tg$ one constant), or do you mean $\tan$? – 2017-02-02
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    Yes,thank you for correcting me! – 2017-02-02
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    @Arthur It's actually $3/2$. – 2017-02-02

3 Answers 3

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I thought that it would be instructive to present a solution that relies only on inequalities that can be obtained without calculus. To that end, we proceed.

In THIS ANSWER, I showed that the tangent function satisfies the inequalities

$$x\le \tan(x)\le \frac{x}{\cos(x)} \tag 1$$

for $0

Using $(1)$, we have

$$\frac{3n}{2n+2}\le n\tan\left(\frac{3}{2n+2}\right)\le \frac{3n}{(2n+2)\cos\left(\frac{3}{2n+2}\right)}$$

whereupon using the squeeze theorem yields the coveted limit

$$\lim_{n\to \infty}n\tan\left(\frac{3n}{2n+2}\right)=\frac32$$

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Hint: Assuming $tg$ means $\tan$, then we have that \begin{align} \lim_{n\rightarrow \infty} n\cdot \tan\frac{3}{2n+2} =&\ \lim_{n\rightarrow \infty} n\cdot \sin\frac{3}{2n+2} \sec \frac{3}{2n+2}\\ =&\ \lim_{n\rightarrow \infty} \frac{3n(2n+2)}{3(2n+2)}\cdot \sin\frac{3}{2n+2} \cdot \sec \frac{3}{2n+2}\\ =&\ \lim_{n\rightarrow \infty} \frac{3n}{(2n+2)}\cdot \frac{\sin\frac{3}{2n+2}}{\frac{3}{2n+2}}\cdot \sec \frac{3}{2n+2}\\ \end{align}

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    Yes, tg stands for tan. Thank you very much for the tip! – 2017-02-02
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Shortly $$\lim _{ n\to \infty } n\cdot \tan { \left( \frac { 3 }{ 2n+2 } \right) } =\lim _{ n\to \infty } \frac { \tan { \left( \frac { 3 }{ 2n+2 } \right) } }{ \frac { 3 }{ 2n+2 } } \cdot \frac { 3n }{ 2n+2 } =\frac { 3 }{ 2 } \\ \\ $$