Problem of limit of power function 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 !

Look this way.
Since $\lim_{x\to 0}\frac{x^2-2x+3}{x^2-3x+2}=\frac{3}{2}$, and $\lim_{x\to 0}\frac{\sin x}{x}=1$, then we have
$$\lim_{x\to 0}\left(\frac{x^2-2x+3}{x^2-3x+2}\right)^\frac{sin x}{x}=\left(\lim_{x\to 0}\frac{x^2-2x+3}{x^2-3x+2}\right)^{\lim_{x\to 0}\frac{\sin x}{x}}=\left(\frac{3}{2}\right)^{(1)}=\frac{3}{2}.$$
