Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 !

share|improve this question
    
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
5  
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

1 Answer 1

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.