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

In summary, the conversation discusses the use of L'Hopital's rule and whether there is a better method to solve a given mathematical problem. One person suggests rewriting the expressions and using double-angle formulas, while another suggests using a Taylor expansion. However, it is noted that the solutions obtained may not match the desired solution.
  • #1
Dustinsfl
2,281
5
L'Hopitals rule here makes it way more complicated. Is there a better method?

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

$$
\lim_{a\to\infty}\left[a^{3/2}(\alpha - \beta -(\sin(\alpha) - \sin(\beta))\right]
$$
 
Physics news on Phys.org
  • #2
Re: Is there better a way than L'Hôpital's rule?

dwsmith said:
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:
  • #3
Re: Is there better a way than L'Hôpital's rule?

I like Serena said:
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 don't think that will arrive at the solution of $\frac{\sqrt{2}}{3}(s^{3/2} - (s-c)^{3/2})$ though.
 
  • #4
Re: Is there better a way than L'Hôpital's rule?

dwsmith said:
I don't 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.
 

FAQ: Is there a better way than L'Hôpital's rule?

What is L'Hôpital's rule and when is it used?

L'Hôpital's rule is a mathematical method used to evaluate limits of indeterminate forms. It states that for two functions f(x) and g(x), if the limit of their quotient approaches 0/0 or ∞/∞, then the limit of f(x)/g(x) is equal to the limit of their derivatives f'(x)/g'(x). This rule is typically used when traditional methods of evaluating limits fail to provide a definitive answer.

Are there any limitations to using L'Hôpital's rule?

Yes, L'Hôpital's rule can only be applied to certain types of indeterminate forms, such as 0/0 and ∞/∞. It also assumes that the functions f(x) and g(x) are differentiable in the given interval.

Is there a better alternative to L'Hôpital's rule?

Yes, there are other methods for evaluating limits of indeterminate forms, such as using algebraic manipulation, substitution, or the squeeze theorem. These methods may be simpler and more intuitive than L'Hôpital's rule, but they may not always provide a solution.

How do I know when to use L'Hôpital's rule?

L'Hôpital's rule should only be used when traditional methods of evaluating limits, such as direct substitution, do not provide a definitive answer. It is also important to check if the given indeterminate form is suitable for applying L'Hôpital's rule.

Can L'Hôpital's rule be used to solve all limit problems?

No, L'Hôpital's rule can only be applied to certain types of indeterminate forms. It is important to understand the limitations of this rule and to use other methods when necessary. Additionally, some limit problems may not have a solution using traditional methods or L'Hôpital's rule.

Similar threads

Back
Top