$\lim_{x \to 0}\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\mu$

Question

I would like to ask kindly for any help to show below limit:

$$\lim_{x \to 0}\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\mu$$

I have tried to use the expansion of $e^x=1+x+x^2/2+x^3/6+\mathcal{O}(x^4)$ but still I can't show it.

Answer 1

Note that

$$\frac{2\mu(e^x(x^2-x+1)-1)-e^x(x^2-2x+2)+2}{x(2\mu(e^x(x-1)+1)-e^x(x-2)-2)}=\frac{[(2\mu-1)x^2-(2\mu-2)x+(2\mu-2)]e^x-(2\mu-2) }{[(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x}=\frac{ [(2\mu-1)x^2-(2\mu-2)x]e^x +(2\mu-2)x +(2\mu-2)e^x-(2\mu-2)x-(2\mu-2)}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=1+(2\mu-2)\frac{ e^x-x-1}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=1+(2\mu-2)\frac{ {e^x-x-1}}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }\to 1+(2\mu-2)\cdot \frac12=\mu$$

indeed

$$\frac{ {e^x-x-1}}{ [(2\mu-1)x^2-(2\mu-2)x]e^x+(2\mu-2)x }=\\=\frac{ {e^x-x-1}}{x^2}\frac{x}{ [(2\mu-1)x-(2\mu-2)]e^x+(2\mu-2) }\to \frac12\cdot 1=\frac12$$

indeed by $e^x=1+x+\frac12x^2+o(x^2)$

$$\frac{ {e^x-x-1}}{x^2}=\frac12 + o(1) \to \frac 12$$

and by $e^x=1+x+o(x)$

$$\frac{x}{ [(2\mu-1)x-(2\mu-2)]e^x+(2\mu-2) }=\frac{x}{ (2\mu-1)x+(2\mu-1)x^2-(2\mu-2)x+o(x)}=\frac{1}{1+(2\mu-1)x+o(1)}\to 1$$

Answer 2

Hint. As regards the denominator, suffices to consider the expansion $e^x=1+x+o(x)$,it $$\begin{align} 2\mu(e^x(x-1)+1)&-e^x(x-2)-2\\&=2\mu((1+x+o(x))(x-1)+1)-(1+x+o(x))(x-2)-2\\&=x+o(x).\end{align}$$ Since the denominator is $x^2+o(x^2)$, for the numerator consider the longer expansion $e^x=1+x+\frac{x^2}{2}+o(x^2)$.