Question Number 131383 by EDWIN88 last updated on 04/Feb/21
![lim_(x→∞) ((x^2 /(x−1)))^(tan ((1/( (√x))))) =?](https://www.tinkutara.com/question/Q131383.png)
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{1}}\right)^{\mathrm{tan}\:\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)} =? \\ $$
Answered by liberty last updated on 04/Feb/21
![L=lim_(x→∞) ((x^2 /(x−1)))^(tan ((1/( (√x))))) L=e^(lim_(x→∞) (tan (1/( (√x))))ln ((x^2 /(x−1)))) = e^(lim_(x→∞) (((2ln x−ln (x−1))/(cot ((1/( (√x)))))))) L = e^(lim_(t→0) (((4ln t−ln (t^2 −1))/(cot ((1/t)))))) = e^(lim_(t→∞) (((2(t^2 −1))/(t(t^2 −1)((1/t^2 ))cosec^2 ((1/t)))))) L=e^(lim_(t→∞) (((sin ((1/t)))/(1/t)))^2 .(((2t^2 −4)/(t^3 −t)))) = e^0 = 1](https://www.tinkutara.com/question/Q131384.png)
$$\mathrm{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{tan}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)} \\ $$$$\mathrm{L}=\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{tan}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)\mathrm{ln}\:\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{1}}\right)} =\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2ln}\:\mathrm{x}−\mathrm{ln}\:\left(\mathrm{x}−\mathrm{1}\right)}{\mathrm{cot}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)}\right)} \\ $$$$\mathrm{L}\:=\:\mathrm{e}^{\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{4ln}\:\mathrm{t}−\mathrm{ln}\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{cot}\:\left(\frac{\mathrm{1}}{\mathrm{t}}\right)}\right)} =\:\mathrm{e}\:^{\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{cosec}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{t}}\right)}\right)} \\ $$$$\mathrm{L}=\mathrm{e}^{\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{t}}\right)}{\frac{\mathrm{1}}{\mathrm{t}}}\right)^{\mathrm{2}} .\left(\frac{\mathrm{2t}^{\mathrm{2}} −\mathrm{4}}{\mathrm{t}^{\mathrm{3}} −\mathrm{t}}\right)} =\:\mathrm{e}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$