How to compute $\lim \limits_{x\to \infty} ((2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3})$ through Taylor series

Question

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 $x=1/t$ 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.

Answer 1

HINT:

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

Answer 2

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 $\infty$.

Answer 3

hint

The square root becomes

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

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

Answer 4

$y = \frac 1x$

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...

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

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