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[SOLVED] Is there a better way than L'Hôpital's rule?

dwsmith

Well-known member
Feb 1, 2012
1,673
L'Hopitals rule here makes it way more complicated. Is there a better method?

$\alpha = 2\arcsin\left(\sqrt{\frac{s}{2a}}\right)$
$\beta = 2\arcsin\left(\sqrt{\frac{s-c}{2a}}\right)$

$$
\lim_{a\to\infty}\left[a^{3/2}(\alpha - \beta -(\sin(\alpha) - \sin(\beta))\right]
$$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Re: Is there better a way than L'Hôpital's rule?

L'Hopitals rule here makes it way more complicated. Is there a better method?

$\alpha = 2\arcsin\left(\sqrt{\frac{s}{2a}}\right)$
$\beta = 2\arcsin\left(\sqrt{\frac{s-c}{2a}}\right)$

$$
\lim_{a\to\infty}\left[a^{3/2}(\alpha - \beta -(\sin(\alpha) - \sin(\beta))\right]
$$
First I would rewrite the expressions for $\alpha, \beta, \sin \alpha, \sin \beta$.
The method to do so, is to draw a right triangle with sides that match your angles.
Combine that with the double-angle-formulas.

You'll get for instance:
$$\quad \alpha = \arccos\left(1-\frac s a\right) \\
\quad \alpha = \arcsin\left(\sqrt{\frac {2as -s^2}{a^2}}\right) \\
\quad \sin \alpha = \sqrt{\frac {2as -s^2}{a^2}}$$

The next step that springs to mind is a Taylor expansion, which is basically a more advanced form of l'Hôpital's rule.
Let's define $A=\sqrt{\dfrac {2as -s^2}{a^2}}$, then for instance, since $\dfrac s a \ll 1$ (near the limit), we have:
$$\quad \alpha = \arcsin(A) = A + \frac {A^3}{6} + \frac{3A^5}{40} + ... \\
\quad \sin \alpha = \sqrt{\frac {2s}{a}}\sqrt{1 - \frac{s}{2a}} = \sqrt{\frac {2s}{a}} \left(1 - \frac 1 2 \cdot \frac{s}{2a} + ...\right)$$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Re: Is there better a way than L'Hôpital's rule?

First I would rewrite the expressions for $\alpha, \beta, \sin \alpha, \sin \beta$.
The method to do so, is to draw a right triangle with sides that match your angles.
Combine that with the double-angle-formulas.

You'll get for instance:
$$\quad \alpha = \arccos\left(1-\frac s a\right) \\
\quad \alpha = \arcsin\left(\sqrt{\frac {2as -s^2}{a^2}}\right) \\
\quad \sin \alpha = \sqrt{\frac {2as -s^2}{a^2}}$$

The next step that springs to mind is a Taylor expansion, which is basically a more advanced form of l'Hôpital's rule.
Let's define $A=\sqrt{\dfrac {2as -s^2}{a^2}}$, then for instance, since $\dfrac s a \ll 1$ (near the limit), we have:
$$\quad \alpha = \arcsin(A) = A + \frac {A^3}{6} + \frac{3A^5}{40} + ... \\
\quad \sin \alpha = \sqrt{\frac {2s}{a}}\sqrt{1 - \frac{s}{2a}} = \sqrt{\frac {2s}{a}} \left(1 - \frac 1 2 \cdot \frac{s}{2a} + ...\right)$$
I dont think that will arrive at the solution of $\frac{\sqrt{2}}{3}(s^{3/2} - (s-c)^{3/2})$ though.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Re: Is there better a way than L'Hôpital's rule?

I dont think that will arrive at the solution of $\frac{\sqrt{2}}{3}(s^{3/2} - (s-c)^{3/2})$ though.
The method I described is guaranteed to work.
Moreover, it is more general than l'Hôpital.