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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 !

  • 0
    Since the base approaches a limit other than $1$ and the exponent approaches a limit, you can just handle those two separately. ${}\qquad{}$2015-12-21

1 Answers 1

3

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

  • 0
    There is no need to SHOUT.2012-11-22