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Evaluation of $$\lim_{x\rightarrow 0}\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} $$

$\bf{My\; Try::}$ Let $$l=\lim_{x\rightarrow 0}\frac{e^{\frac{\ln(1+2x+3x^2)}{x}}-e^{\frac{\ln(1+2x-3x^2)}{x}}}{x}$$

Using $$\bullet \; \frac{\ln(1+x)}{x}=x-\frac{x^2}{2}+\frac{x^3}{3}-.....\infty$$

But I am not Getting answer.

Now How can I solve after that, Help me

Thanks

Did
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juantheron
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3 Answers3

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As usual this limit can also be evaluated without the use of L'Hospital's Rule and Taylor's series just by applying standard limits combined with the use of algebra of limits. We have \begin{align} L &= \lim_{x \to 0}\frac{(1 + 2x + 3x^{2})^{1/x} - (1 + 2x - 3x^{2})^{1/x}}{x}\notag\\ &= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x + 3x^{2})}{x}\right) - \exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{x}\right)}{x}\notag\\ &= \lim_{x \to 0}\exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{x}\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + 2x + 3x^{2}) - \log(1 + 2x - 3x^{2})}{x}\right) - 1}{x}\notag\\ &= \lim_{x \to 0}\exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{2x - 3x^{2}}\cdot(2 - 3x)\right)\cdot\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right)\right) - 1}{x}\notag\\ &= \exp(1\cdot 2)\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right)\right) - 1}{x}\notag\\ &= \exp(2)\lim_{z \to 0}\frac{\exp(z) - 1}{z}\cdot\lim_{x \to 0}\dfrac{z}{x}\notag\\ &= \exp(2)\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{6x^{2}}{1 + 2x - 3x^{2}}\right)}{x^{2}}\notag\\ &= \exp(2)\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{6x^{2}}{1 + 2x - 3x^{2}}\right)}{\dfrac{6x^{2}}{1 + 2x - 3x^{2}}}\cdot\frac{6}{1 + 2x - 3x^{2}}\notag\\ &= e^{2}\cdot 1\cdot 6 = 6e^{2}\notag \end{align} We have used the standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1 = \lim_{x \to 0}\frac{e^{x} - 1}{x}$$ and the fact that $$z = \frac{1}{x}\log\left(\frac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right) = \frac{1}{x}\log\left(1 + \frac{6x^{2}}{1 + 2x - 3x^{2}}\right)\to 0$$ as $x \to 0$ (which easily follows from the standard limits mentioned above).

Paramanand Singh
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Consider $$A=\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} =\frac{B^{\frac{1}{x}}-C^{\frac{1}{x}}}x$$ using $$B=(1+2x+3x^2) \qquad , \qquad C=(1+2x-3x^2)$$ So $$\log(B^{\frac{1}{x}})=\frac{1}{x}\log(B)$$ and now, using Taylor series $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ replace $y$ by $(2x+3x^2)$ to get $$\log(B)=2 x+x^2-\frac{10 x^3}{3}+O\left(x^4\right)$$ which gives $$\log(B^{\frac{1}{x}})=2+x-\frac{10 x^2}{3}+O\left(x^3\right)$$ Now, using Taylor again, $$B^{\frac{1}{x}}=e^{\log(B^{\frac{1}{x}})}=e^2+e^2 x-\frac{17 e^2 x^2}{6}+O\left(x^3\right)$$ Doing the same with the second term, you should arrive to $$C^{\frac{1}{x}}=e^2-5 e^2 x+\frac{127 e^2 x^2}{6}+O\left(x^3\right)$$ All of that makes $$A=6 e^2-24 e^2 x+O\left(x^2\right)$$ which shows the limit and how it is approached.

Claude Leibovici
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  • Was waiting for your answer just to check what is the next term after $6e^{2}$ but you went even further to $127e^{2}x^{2}/6$ :):) +1 – Paramanand Singh May 20 '16 at 09:23
  • @ParamanandSingh. As I alredy said a few times, I felt in love with Taylor series more than 60 years ago and I always look at the asymptotics (this is the way I learnt this stuff).Have a look at http://matheducators.stackexchange.com/questions/8339/teaching-limits-and-asymptotics-at-the-same-time – Claude Leibovici May 20 '16 at 09:27
  • Thanks for the link to nice discussion on matheducators. Maybe I should also ask some stuff occasionally in that forum, although like you I am also not a math educator myself. – Paramanand Singh May 20 '16 at 09:33
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Hint: $e^{1+t} - e^{1-t} \sim 2t$ as t tends to 0

Alex
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