I have a problem :
Find
$$\lim_{x\to 0}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)^{\dfrac{\sin x}{x}}$$
Here is my argument :
$$\lim_{x\rightarrow 0}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)^{\dfrac{\sin x}{x}}= \lim_{x\rightarrow 0}e^{\ln\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)\dfrac{\sin x}{x}}$$
On the other hand,
$$\lim_{x\rightarrow 0}\text{ln}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)=\ln\left(\lim_{x\rightarrow 0}\dfrac{x^2-2x+3}{x^2-3x+2}\right)=\text{ln}\dfrac{3}{2}$$
and
$$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$
therefore
$$\lim_{x\rightarrow 0}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)^{\dfrac{\sin x}{x}}=e^{\ln\frac{3}{2}}=\dfrac{3}{2}$$
Am I wrong ? If I am wrong, please show me how to do this problem.
Thanks !