Hi,

I'm not sure how to find out the answer to this qustion so any help would be great. I have a hunch thatKmay be -1 andlmay be +/-4.

Thanks, DLL.

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- May 27th 2009, 09:05 PMDaddy_Long_LegsFinding the Values Using a Limit
Hi,

I'm not sure how to find out the answer to this qustion so any help would be great. I have a hunch that**K**may be -1 and**l**may be +/-4.

Thanks, DLL. - May 27th 2009, 09:14 PMSpec
Try replacing $\displaystyle \cos lx$ with the Maclaurin series for it.

- May 28th 2009, 02:48 AMDaddy_Long_Legs
Hi,

I tried doing the Maclaurin series but I'm not sure where to go from there. I can't see where finding the values a**k**and**l**comes into it.

Thanks, DLL - May 28th 2009, 03:58 AMHallsofIvy
The denominator is going to 0 so, in order to have any finite limit, the numerator must also go to 0. Since cos(x) goes to 1 as x goes to 0, yes, k must be -1.

You then have $\displaystyle \frac{-1+ cos(lx)}{x^2}$. I would suggest using L'Hopital's rule: differentiating numerator and denominator separately, we get $\displaystyle \frac{-l sin(lx)}{2x}= -\frac{l^2}{2}\frac{sin(lx)}{lx}$ where I have multiplied numerator and denominator by l to get "sin(z)/z" as z goes to 0. - May 28th 2009, 07:34 AMSpec
Maclaurin series: $\displaystyle \cos lx=1-\frac{(lx)^2}{2}+O(x^4)$

Put that into the limit and you get $\displaystyle \lim_{x \to 0}\frac{k+1-\frac{(lx)^2}{2}+O(x^4)}{x^2}=\frac{k+1}{x^2}-\frac{\frac{(lx)^2}{2}+O(x^4)}{x^2}=-4$

We can see that $\displaystyle k=-1$ since otherwise the limit wouldn't be finite.

With that out of the way, we have $\displaystyle \lim_{x \to 0}-\frac{\frac{(lx)^2}{2}+O(x^4)}{x^2}=\lim_{x \to 0}\left(-\frac{l^2}{2}+O(x^2)\right)=-4 \Longleftrightarrow l^2=8$ since $\displaystyle \lim_{x \to 0} O(x^2)=0$ - May 29th 2009, 04:48 AMDaddy_Long_Legs
HallsofIvy (Wait)

I tried doing the problem the way you sugested but I'm not sure how to find the value of**l**after you get to "sin(z)/z" as z goes to 0. I reasonably certain the value of**l**is -/+4. I just can't seem to find how to find it.

Thanks. - May 29th 2009, 04:55 AMmr fantastic
If you're going to use l'Hopital's rule once, you may as well use it twice.

Using it the second time gives $\displaystyle \lim_{x \rightarrow 0} \frac{-l^2 \cos (lx)}{2}$. So $\displaystyle \frac{-l^2}{2} = -4$ ....

By the way, you're expected to know that $\displaystyle \lim_{x \rightarrow 0} \frac{\sin z}{z} = 1$ ....