How do I evaluate the limit $$ \lim_{x \to 0}\frac{e^{x^2-x} -1 + x - \alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}$$

depending on $\alpha, $using Taylor series? I know I consider

$e^t = 1 +t +\frac{t^2}{2} + o(t^2)$

$ \log(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} + o(t^3)$

$\cosh(t) = 1+ \frac{t^2}{2} + \frac{t^4}{24} + o(t^4)$

$\sin(t) = t - \frac{t^3}{6} + \frac{t^5}{120} + o(t^5)$

and substitute. I know why it is possibile and to justify this method. Though, I am not able to solve this limit... Thanks!

  • 1
    $\begingroup$ So why don't you show us plugging stuff in. I can already see that the $1's$ cancel due to e-power and cosh(2x) terms. Next, you may be able to cancel common $x$ terms as a result of the constants gone. $\endgroup$ – imranfat Feb 2 '18 at 16:50

Note that

$$\begin{cases}x^3 \cdot x \sin (1/x^2)\to 0 \implies x^4\sin (1/x^2)=o(x^3)\\\\ x^3 \cdot x\log x\to 0\implies x^4\log x=o(x^3)\\\\ e^{x^2-x} -1 =x^2-x+\frac{(x^2-x)^2}{2}+o(x^2)=-x+\frac32x^2+o(x^2)\\\\ \cosh 2x=1+2x^2+o(x^2)\end{cases}$$


$$\frac{e^{x^2-x} -1 -\alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)} =\frac{-x+\frac32x^2+o(x^2) - \alpha x^2 + o(x^3) }{1+2x^2+o(x^2) -1 + o(x^3)} =\frac{-x+\left(\frac32-\alpha \right)x^2 +o(x^2) }{2x^2+o(x^2)}=\frac{-\frac1x+\left(\frac32-\alpha \right) +o(1) }{2+o(1)}\to\pm\infty $$

therefore $$\lim_{x \to 0}\frac{e^{x^2-x} -1 - \alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}$$

does not exist.

  • $\begingroup$ Thanks! I forgot the term $+x$ in the numerator, so it's just $-\alpha$. By the way, I can do it solve it on my own now. Just a doubt, when you expand $e^{x^2-x}$, is it not $o(x^4)$? Because if I substitute I get $o((x^2-x)^2) = o(x^4)$. $\endgroup$ – Francesco Gemma Feb 3 '18 at 15:15
  • $\begingroup$ You are welcome! How you can see now, $\alpha$ term is not important here for that I let $\pm$ also if I presumed it was a typo. Note that $o((x^2-x)^2)=o(x^4-2x^3+x^2)=o(x^2)$. If you have doubts on it refer to the definition $o((x^2-x)^2)=(x^2-x)^2\cdot \omega ((x^2-x)^2)$. The key point in this limit is just to recognize, by the definition, that $x^4\log x$ and $x^4 \sin(1/x^2)$ are both $o(x^3)$ otherwise since you can't expand them it should be difficult conclude. $\endgroup$ – user Feb 3 '18 at 15:21

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