Menu Close

lim-n-tan-pi-4-1-n-n-t-1-n-lim-t-0-tan-pi-4-t-1-t-lim-t-0-tan-pi-4-t-1-1-t-lim-t-0-1-tant-1-tant-1-1-t-lim-t-0-2tant-1-tant-1-t-2-An




Question Number 220473 by mehdee7396 last updated on 13/May/25
lim_(n→∞) (tan((π/4)+(1/n)))^n      →^(t=(1/n))   =lim_(t→0) [tan((π/4)+t)]^(1/t)   ⇒lim_(t→0) [tan((π/4)+t)−1]×(1/t)  =lim_(t→0) (((1+tant)/(1−tant))−1)×(1/t)  =lim_(t→0) (((2tant)/(1−tant)))×(1/t)=2  ⇒Ans=e^2   ✓
$${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2} \\ $$$$\Rightarrow{Ans}={e}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *