# Help me with a limit problem please

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Posts: 5

Member No.: 295

Joined: 26-June 04

2004-06-27

Hello there.

My name is Hans Lindroth. I've just started to study maths at the university. Now I've encountered a problem about limits. It goes like this: "In a circle with radius 'R' a chord is drawn, the length of which is 'L'. Let 'H' be the distance between the center of the chord and the center of the smaller arc being cut off the circle by the chord. Decide the limit:

lim H/(L^2)".

L->0

I've tried by finding an expression for "H" in terms of "L".

(which is "1/8(R^2)") is given in terms of 'R' (which means I must have lost myself).

Thanks

Hans Lindroth.

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Haven't found a solution yet, but I'll post what I have. If you construct your chord and then run a radius to one of the edges of the chord and through the centre of it, you get:

$\left( \frac{L}{2} \right)^2 +(R-H)^2 = R^2$.

Just by looking at the diagram, it's pretty obvious that as L->0 so does H, but I get the limit coming out as $\frac{1}{8R-1}$?

I dunno. Can't be bothered beating the solution out atm.

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I get 1/8R. This probably the most obvious way to do it.

if u look at the circle, you can write L and H in terms of R and x . we dont have to worry what happens to R cos its constant, and x is the only variable.

so we have

$\frac{L}{2}=Rcosx$

$L=2Rcosx$

and

$K=Rsinx$

therefore

$H=R-Rsinx$

$H=R(1-sinx)$

so

$\frac{H}{L^2}=\frac{R(1-sinx)}{4R^{2}\cos^{2}x}$

$=\frac{(1-sinx)}{4R(1-\sin^{2}x)}$

$=\frac{(1-sinx)}{4R(1-sinx)(1+sinx)}$

$=\frac{1}{4R(1+sinx)}$

Letting L tend to 0 is the same as letting x tend to Pi/2

therefore

limit of $=\frac{1}{4R(1+sinx)}$ as x tends to Pi/2 , by ALG

$=\frac{1}{4R(1+1)}$

$=\frac{1}{8R}$

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Nicely done; I can't see where I got the -1 from on the bottom of my fraction. I'll have a look later.

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