Question Number 220380 by MrGaster last updated on 12/May/25
![lim_(n→∞) tan[(π/4)+(1/n)]^n =?](https://www.tinkutara.com/question/Q220380.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}\left[\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right]^{{n}} =? \\ $$
Answered by SdC355 last updated on 12/May/25
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\mathrm{tan}\left(\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \right)=\mathrm{tan}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)^{{n}} =\mathrm{tan}\left({e}^{{n}\centerdot\mathrm{ln}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{tan}\left(-\right)=\mathrm{tan}\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:-\right)=\mathrm{tan}\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{n}\centerdot\mathrm{ln}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{n}\centerdot\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)} =\mathrm{0} \\ $$$$\mathrm{tan}\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Answered by mr W last updated on 12/May/25
![=lim_(n→∞) tan {((π/4))^n [(1+(4/(nπ)))^((nπ)/4) ]^(4/π) } =tan {0×e^(4/π) } =0](https://www.tinkutara.com/question/Q220387.png)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}\:\left\{\left(\frac{\pi}{\mathrm{4}}\right)^{{n}} \left[\left(\mathrm{1}+\frac{\mathrm{4}}{{n}\pi}\right)^{\frac{{n}\pi}{\mathrm{4}}} \right]^{\frac{\mathrm{4}}{\pi}} \right\} \\ $$$$=\mathrm{tan}\:\left\{\mathrm{0}×{e}^{\frac{\mathrm{4}}{\pi}} \right\} \\ $$$$=\mathrm{0} \\ $$
Commented by mr W last updated on 12/May/25
Answered by MrGaster last updated on 12/May/25
Commented by mr W last updated on 12/May/25
![but the question is lim_(n→∞) tan ((π/4)+(1/n))^n ,not lim_(n→∞) [tan ((π/4)+(1/n))]^n .](https://www.tinkutara.com/question/Q220417.png)
$${but}\:{the}\:{question}\:{is} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:,{not} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right]^{{n}} . \\ $$
Commented by MrGaster last updated on 12/May/25
misunderstood the problem, the original problem and expression is actually the limit of [tan(pi/4 +1/n)]^n.
Commented by mr W last updated on 12/May/25
$${then}\:{e}^{\mathrm{2}} \:{is}\:{correct}. \\ $$
Commented by MrGaster last updated on 12/May/25
yes
Commented by MrGaster last updated on 12/May/25
Commented by mehdee7396 last updated on 13/May/25
![lim_(n→∞) [tan((π/4)+(1/n))−1]×n lim_(n→∞) [((1+tan(1/n))/(1−tan(1/n)))−1]×n lim_(n→∞) [((2tan(1/n))/(1+tan(1/n)))]×n lim_(t→0) [((2tant)/t)]×(1/(1+tant))=2 ⇒Ans=e^2 ✓](https://www.tinkutara.com/question/Q220472.png)
$${lim}_{{n}\rightarrow\infty} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)−\mathrm{1}\right]×{n} \\ $$$${lim}_{{n}\rightarrow\infty} \left[\frac{\mathrm{1}+{tan}\frac{\mathrm{1}}{{n}}}{\mathrm{1}−{tan}\frac{\mathrm{1}}{{n}}}−\mathrm{1}\right]×{n} \\ $$$${lim}_{{n}\rightarrow\infty} \left[\frac{\mathrm{2}{tan}\frac{\mathrm{1}}{{n}}}{\mathrm{1}+{tan}\frac{\mathrm{1}}{{n}}}\right]×{n} \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} \left[\frac{\mathrm{2}{tant}}{{t}}\right]×\frac{\mathrm{1}}{\mathrm{1}+{tant}}=\mathrm{2} \\ $$$$\Rightarrow{Ans}={e}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$