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$$\underset{x\to \infty}{\lim} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$

My Attempt: $$L = \underset{t\to 0}{\lim} \frac{\left( \left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2\right)} {t^2}$$ I Now have a $\frac 00$ form that I could use L'Hopital rule with, but I don't want to differentiate the ugly looking function in the numerator. Is there an easier way to solve these kinds of problems? Maybe a taylor series expansion for $(1+t)^{\frac 1t}, t \to 0$ forms would come in handy here and I could just subtract the $e^2$ from the resulting expansion.

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    $\begingroup$ I like the idea of a series expansion. If you find the series expansion of $\frac1t\log\frac{t+1}{t-1}$ (to the $t^3$ term, say), then you can exponentiate that (using the series expansion of $e^u$ to the $u^3$ term, say) to get the expansion of $(\frac{t+1}{t-1})^t$. $\endgroup$ Feb 29 '20 at 4:29
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Let $y=\left( \frac{t+1}{t-1} \right)^{\frac 1t}$. Then,

$$\ln y = \frac1t \ln \frac{t+1}{t-1} = \frac1t [\ln (1+t)-\ln(1-t)]$$ $$= \frac1t [(t-\frac12t^2+\frac13t^3) - (t+\frac12t^2-\frac13t^3) +O(t^5)] =2+\frac23t^2+O(t^4)$$

and

$$\left( \frac{t+1}{t-1} \right)^{\frac 1t}=y= e^{2+\frac23t^2+O(t^4)} = e^2(1+\frac23t^2+O(t^4))$$

Thus, $$L = \underset{t\to 0}{\lim} \frac{\left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2} {t^2} = \underset{t\to 0}{\lim} \frac{e^2(1+\frac23t^2+O(t^4))-e^2}{t^2} = \underset{t\to 0}{\lim} \left(\frac23e^2+O(t^2)\right)=\frac23e^2$$

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$$ \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$

Start with $$y=\left( \frac{x+1}{x-1} \right)^x\implies \log(y)=x\log\left( \frac{x+1}{x-1} \right)=x\log\left(1+ \frac{2}{x-1} \right)$$ Now, by Taylor $$\log(y)=x\left(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+O\left(\frac{1}{x^7}\right)\right)={2}+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$ $$y=e^{\log(y)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45 x^4}+O\left(\frac{1}{x^6}\right)$$ $$(y-e^2)x^2=\frac{2 e^2}{3}+\frac{28 e^2}{45 x^2}+O\left(\frac{1}{x^4}\right)$$

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First,you made a mistake.$L$ should be $$ \lim_{t\to 0}\dfrac{\left(\dfrac{1+t}{1-t}\right)^{\dfrac{1}{t}}-e^2}{t^2}$$

Then,we could use equivalent substitution to compute the limit as follow. $$ln(1+t)=t-\dfrac{t^2}{2}+\dfrac{t^3}{3}+o(t^4)$$ $$\dfrac{1}{t}(ln(1+t)-ln(1-t))=2+\dfrac{2t^2}{3}+o(t^2)$$ $$\lim_{t\to 0}\dfrac{1}{t}ln\left( \dfrac{1+t}{1-t}\right)-2=0$$ $$L=\lim_{t\to 0}\dfrac{( e^{\dfrac{1}{t}ln\left( \dfrac{1+t}{1-t}\right)-2}-1) e^2}{t^2}=\lim_{t\to 0}(\dfrac{1}{t}ln\left(\dfrac{1+t}{1-t}\right)-2)\frac{e^2}{t^2}$$ $$=\lim_{t\to 0}\frac{e^2}{t^2}(\dfrac{2t^2}{3})=\dfrac{2e^2}{3} $$

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  • $\begingroup$ Thanks.you are right. $\endgroup$
    – Mr.xue
    Feb 29 '20 at 8:37
  • $\begingroup$ No problem. Also, to format functions properly type a slash before the function. For example, "\ln(x)" will give $\ln(x)$. The same goes for other functions. $\endgroup$
    – KM101
    Feb 29 '20 at 8:38
  • $\begingroup$ OK,I know it.I will be careful later. $\endgroup$
    – Mr.xue
    Feb 29 '20 at 8:43

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