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I am trying to compute the below limit through Taylor series: $\lim \limits_{x\to \infty} ((2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3})$

What I have already tried is first of all change the variable x to $x=\frac{1}{t}$ and the limit to $t$ limits to $0$, so I am able to use Maclaurin series. After that I change $e$ to exponent polynomial up to $t=6$ + $O(X^6)$ however, I don`t know what can I do with square root.

Red Five
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John D
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4 Answers4

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HINT:

$$\sqrt{x^6+3}=\frac1{t^3}\left(1+3t^6\right)^{1/2}=\frac1{t^3}+\frac{\frac12\cdot3t^6}{t^3}-\frac{\frac1{2\cdot4}\cdot(3t^6)^2}{t^3}+\frac{\frac{1\cdot3}{2\cdot4\cdot6}\cdot(3t^6)^3}{t^3}-\cdots\\(2x^3-2x^2+x)e^{1/x}=\left(\frac2{t^3}-\frac2{t^2}+\frac1t\right)\left(1+t+\frac{t^2}{2\cdot1}+\frac{t^3}{3\cdot2\cdot1}+\cdots\right)$$

TheSimpliFire
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  • Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it? – John D Dec 28 '18 at 20:00
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    @JohnD $$\sqrt{x^6+3}=\sqrt{\frac1{t^6}+3}=\sqrt{\frac1{t^6}(1+3t^6)}=\frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+\frac12(3t^6)-\frac1{2\cdot4}(3t^6)^2+\cdots$$ – TheSimpliFire Dec 28 '18 at 20:04
  • Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2(1+3t^6)^-1/2 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated! – John D Dec 28 '18 at 20:43
  • Please see here and here. The binomial series with $\alpha=1/2$ is a particular application of Taylor/Maclaurin series. – TheSimpliFire Dec 28 '18 at 20:58
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I think that you don't have to go that far. You know that (for large values of $x$): $$ e^{1/x}=\sum_{k=0}^{\infty}{\frac{1}{k!x^k}}\geq1 $$ so that: $$ (2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3} \geq (2x^3-2x^2+x)-\sqrt{x^6+3}\geq\frac{x^3}{2} $$ Thus, the limit goes to $\infty$.

DudeMan
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hint

The square root becomes

$$\frac{\sqrt{1+3t^6}}{|t^3|}=$$

$$\frac{1}{|t^3|}\Bigl(1+\frac{3t^6}{2}-\frac 98t^{12}+t^{12}\epsilon(t)\Bigr)$$

hamam_Abdallah
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$y = \frac 1x$

then we have

$\lim_\limits{y\to 0^+} \frac {(2 -2y+ y^2)e^y - \sqrt {1+3y^6}}{y^3}$

Now if you want to do a Taylor expansion...

$\lim_\limits{y\to 0^+} \frac {(2 -2y+ y^2)(1+y+\frac 12 y^2+\cdots) - (1+\frac 32 y^6 - \cdots )}{y^3}$

$\lim_\limits{y\to 0^+} \frac {1 + O(y)}{y^3} = \infty$

Doug M
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