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I have a problem :


$$\lim_{x\to 0}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)^{\dfrac{\sin x}{x}}$$

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,

$$\lim_{x\rightarrow 0}\text{ln}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)=\ln\left(\lim_{x\rightarrow 0}\dfrac{x^2-2x+3}{x^2-3x+2}\right)=\text{ln}\dfrac{3}{2}$$


$$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$


$$\lim_{x\rightarrow 0}\left(\dfrac{x^2-2x+3}{x^2-3x+2}\right)^{\dfrac{\sin x}{x}}=e^{\ln\frac{3}{2}}=\dfrac{3}{2}$$

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 the base approaches a limit other than $1$ and the exponent approaches a limit, you can just handle those two separately. ${}\qquad{}$ – Michael Hardy Dec 21 '15 at 17:08

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

share|cite|improve this answer
There is no need to SHOUT. – Gerry Myerson Nov 22 '12 at 0:35

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