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If $\displaystyle\lim_{x\to \infty} \left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x} = e^2$, find $a,b\in \mathbb{R}$

I tried this by converting the expression given inside the brackets (denote by $f(x)$) into $e^\left({\displaystyle\lim_{x\to \infty}{2x}{\ln(f(x))}}\right)$ and then evaluating the power with $2$. But I didn't really get the correct solution.

I found a solution here but I couldn't understand how they did. I feel that they were along the same lines as I except that either of us did something wrong.

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  • $\begingroup$ Try using the fact that: $\displaystyle\lim_{x \rightarrow \infty} \left(1+ \frac{2}{x}\right)^x = e^2$. $\endgroup$ – Mateusz Eggink May 3 '18 at 20:22
  • $\begingroup$ use the expansion of $\ln(1+\frac ax+\frac b{x^2})=\frac ax+\frac{b-\frac {a^2}2}{x^2}$. $\endgroup$ – zwim May 3 '18 at 20:25
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 3 '18 at 20:46
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Note that we have

$$\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}=\left[\left(1+\frac{ax+b}{x^2}\right)^{\frac{x^2}{ax+b}}\right]^\frac{2ax^2+2xb}{x^2}\to e^{2}$$

when $a=1$ for every $b$.

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Set $$1+\frac{a}{x}+\frac{b}{{{x}^{2}}}=\left( 1+\frac{\alpha }{x} \right)\left( 1+\frac{\beta }{x} \right)=1+\frac{\alpha +\beta }{x}+\frac{\alpha \beta }{{{x}^{2}}}$$ So $$a=\alpha +\beta \ \ \text{ and }\ \ b=\alpha \beta ,$$ Now $$\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}^{2x}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{\alpha }{x} \right)}^{2x}}{{\left( 1+\frac{\beta }{x} \right)}^{2x}}={{e}^{2\alpha }}{{e}^{2\beta }}={{e}^{2\left( \alpha +\beta \right)}}={{e}^{2a}}$$ Hence ${{e}^{2a}}={{e}^{2}}\Rightarrow a=1$ and the valve of $b$ is arbitrary.

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You can use the Taylor series of $ln(1+u) = u-\frac{u^2}{2}+\frac{u^3}{3} ...$

Thus, letting $u=\frac{a}{x}+\frac{b}{x^2}$ we have
$ln(1+\frac{a}{x}+\frac{b}{x^2})^{2x} = 2x\left( (\frac{a}{x}+\frac{b}{x^2})-(\frac{a}{x}+\frac{b}{x^2})^2 ...\right) = 2a+\frac{2b-a^2}{x} ...$

All further terms will have even higher powers of $x$ in the denominator, so as $x \rightarrow \infty$, they go to zero. Taking the log of the other side, you get $2a = 2e$, so $a = 1$ and b is free.

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we know that

$$\ln (1+X)\sim X (X\to 0)$$

thus

$$\ln (1+\frac {a}{x}+\frac{b}{x^2})=$$ $$\ln\Bigl(1+\frac {a}{x}(1+\frac {b}{ax})\Bigr)$$ $$\sim \frac{a}{x} \; (x\to +\infty)$$

the limit is surely $$e^{2a} $$

So $a=1$ and $b $ is arbitrary.

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With $f(x)=1+\frac{a}{x}+\frac{b}{x^2}$ you have when $ x \to + \infty$: $$\ln(f(x))= \ln \left(1+\frac{a}{x}+o\left(\frac{1}{x} \right) \right)=\frac{a}{x}+o\left(\frac{1}{x}\right)$$ so: $$ 2x \ln(f(x))= 2a+o(1)$$ i.e: $$\lim_{x \to + \infty} 2x \ln(f(x))=2a$$

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  • $\begingroup$ You have to show that b can be ignored. $\endgroup$ – marty cohen May 3 '18 at 23:20
  • $\begingroup$ $b$ is ignored as $b/x^2=o(1/x)$. $\endgroup$ – Delta-u May 4 '18 at 6:17

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