# 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 !

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It is correct. Hopefully you know how to justify each step, though. –  DonAntonio Nov 14 '12 at 2:32
Looks OK to me. –  Gerry Myerson Nov 14 '12 at 2:32
Excellent work! –  Cameron Buie Nov 14 '12 at 2:38
In addition to the wonderful answers/validations you received, you could always use a CAS to verify your work. For example, at WolframAlpha, you can type: Limit[((x^2-2*x+3)/(x^2-3*x+2))^(Sin[x]/x),x->0] –  Amzoti Nov 21 '12 at 18:22

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 $$\lim_{x\to 0}\left(\frac{x^2-2x+3}{x^2-3x+2}\right)^\frac{sin x}{x}=\left(\lim_{x\to 0}\frac{x^2-2x+3}{x^2-3x+2}\right)^{\lim_{x\to 0}\frac{\sin x}{x}}=\left(\frac{3}{2}\right)^{(1)}=\frac{3}{2}.$$