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For $a \geq 0 $, the following limit

$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$

can be computed by applying L'Hopital rule as follows

$$L = \exp \left(\lim_{x \rightarrow \infty} \frac{\log \left(1+\dfrac{1}{x} \right)}{\frac{1}{\sqrt{ax^2 +bx+c}}}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{\frac{-1}{x(x+1)}}{\frac{-(2ax+b)}{2(ax^2 +bx+c)^{3/2}}}\right) =$$ $$ = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2(ax^2 +bx+c)^{3/2}}{x(x+1)(2ax+b)}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2\left(a+ \frac{b}{x}+\frac{c}{x^2}\right)^{3/2}}{1\cdot\left(1+\frac{1}{x}\right)\left(2a+\frac{b}{x}\right)}\right) = \exp(\sqrt{a}).$$

I am wondering if exists another method to compute this limit.

Thanks for any hint!

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    $\begingroup$ One can use the conjugate quantity, that is, note that $$\left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}=\left(1+\frac{1}{x} \right)^{u(x)}\left(\left(1+\frac{1}{x} \right)^x\right)^{\sqrt{a}}$$ with $$u(x)=\sqrt{ax^2 +bx+c}-x\sqrt{a}=\frac{bx+c}{x\sqrt{a}+\sqrt{ax^2 +bx+c}}\to v=\frac{b}{2\sqrt{a}}$$ hence $$\left(1+\frac{1}{x} \right)^{u(x)}\to(1+0)^v=1$$ while $$\left(\left(1+\frac{1}{x} \right)^x\right)^{\sqrt{a}}\to\left(e\right)^{\sqrt{a}}$$ $\endgroup$ – Did Jan 17 '17 at 10:38
  • $\begingroup$ Thanks @Did! Nice answer! :) $\endgroup$ – Alex Silva Jan 17 '17 at 11:11
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I would start with

$$\left(1+\frac1x\right)^{\sqrt{ax^2+bx+c}} = \left(1+\frac1x\right)^{\sqrt a x \cdot\sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}} =\left(\left(1 + \frac1x\right)^x\right)^{\sqrt{a}\cdot \sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}}$$

Now, use the fact that, if $f$ and $g$ are continuous and all limits exist, $$\lim_{x\to\infty}f(x)^{g(x)} = \left(\lim_{x\to\infty}f(x)\right)^{\lim_{x\to\infty} g(x)}$$

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  • $\begingroup$ Why is your last statement true about continuous functions? Could you provide an explanation of why that doesn't work for $\left(1+\frac{1}{x}\right)^x$ like many seem to think? $\endgroup$ – user12345 Jan 17 '17 at 13:55
  • $\begingroup$ @anonymaker000010001 Because for $f(x)=x$, the limit $\lim_{x\to\infty} f(x)$ doesn't exist. I specifically said "and all limits exist" $\endgroup$ – 5xum Jan 17 '17 at 13:56
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Hint

$$A=\left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}\implies \log(A)={\sqrt{ax^2 +bx+c}}\log\left(1+\dfrac{1}{x} \right)$$ Now, use Taylor series $$\log\left(1+\dfrac{1}{x} \right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ and you will arrive to the result (not only the limit but also how it is approached).

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  • $\begingroup$ @AlexSilva. You are very welcome ! I must confess once more my passion for Taylor series. They are so simple and useful. If you noticed in may naswer, I gave you the way to get more than the limit. This was on purpose; in your case, you get the asymptote of the function too. $\endgroup$ – Claude Leibovici Jan 17 '17 at 11:17

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