4
$\begingroup$

$$\lim_{x \to \frac{\pi}{2}}\frac{b(1-\sin x) }{(\pi-2x)^2}$$

I had been solving questions like these using L'Hôpital's rule since weeks. But today, a day before the examination, I got to know that its usage has been 'banned', since we were never officially taught L'Hôpital's rule.

Now I am at a loss how to solve questions I breezed through previously. While it is not possible to learn in 18 hours methods that would cover all kinds of problems, I am hoping I can pick up enough of them to salvage the examination tomorrow.

It has been hinted to me that the above could be solved using trigonometric techniques, but I'm at a loss as to how.

$\endgroup$
6
$\begingroup$

That's a good thing L'Hospital's rule has been banned. If it is not well applied, it can lead to errors, and when it works, using Taylor's formula at order $1$ is logically equivalent. Very often, using equivalents is the shortest way to compute a limit.

That said, use substitution: set $x=\dfrac\pi2-h$; $h\to 0$ if $x\to\dfrac\pi2$. Then $$\frac{b(1-\sin x)}{(\pi-2x)^2}=\frac{b(1-\cos h)}{4h^2}$$ Now it is a standard limit that $$\lim_{h\to 0}\frac{1-\cos h}{h^2}=\lim_{h\to 0}\frac{1-\cos^2 h}{h^2(1+\cos h)}=\lim_{h\to 0}\Bigl(\frac{\sin h}h\Bigr)^2\frac1{(1+\cos h)}=\frac12.$$ Thus the limit in question is equal to $\color{red}{\dfrac b8}.$

$\endgroup$
  • 3
    $\begingroup$ One of the commonest errors is to apply the rule repeatedly to solve an indetermination and forget it is valid only in case of indetermination, going one step too far. (a caricature would be to obtain at some step $\frac x1$, proceed to $\frac 10$ and conclude to an infinite limit!) $\endgroup$ – Bernard Sep 27 '15 at 13:51
  • 8
    $\begingroup$ But by that logic the incorrect application of any theorem can lead to errors, so all theorems must be avoided. $\endgroup$ – R R Sep 27 '15 at 13:59
  • 1
    $\begingroup$ The difference is it is not a real theorem: as I said, it is equivalent to Taylor at order $1$. With Taylor, you won't have the problem. So why use an error-prone rule that can be replaced with a more general formula, that does not have the same pitfall? I don't ay it never must be used, but very often students try to compute improbable derivatives just to apply their beloved L'Hospital's rule… $\endgroup$ – Bernard Sep 27 '15 at 14:11
  • 2
    $\begingroup$ I don't think it is that confusing: it's italic, there is a lonely $h^2$ in the denominator, and it is a traditional notation for small increments of the variable. What do you think of $\ln n$ then? $\endgroup$ – Bernard Sep 27 '15 at 22:11
  • 1
    $\begingroup$ Of course it's an extension to indeterminate forms other than $0/0$ and other special cases, but ultimately, it is really nothing more. In any case I've never seen a limit which could not be solved with Taylors formula, but was solved with L'Hospital. $\endgroup$ – Bernard Sep 27 '15 at 22:58
4
$\begingroup$

$$\begin{align}\lim_{x\to \pi/2}\frac{b(1-\sin x)}{(\pi-2x)^2}&=\lim_{x\to \pi/2}\frac{b(1-\cos(x-\frac{\pi}{2}))}{(\pi -2x)^2}\\\\&=\lim_{x\to\pi/2}\frac{b\sin^2(x-\frac{\pi}{2})}{4(x-\frac{\pi}{2})^2(1+\cos(x-\frac{\pi}{2}))}\\\\&=\lim_{x\to\pi/2}\frac{b}{4}\cdot\frac{1}{1+\cos(x-\frac{\pi}{2})}\cdot\left(\frac{\sin(x-\frac{\pi}{2})}{x-\frac{\pi}{2}}\right)^2\\\\&=\frac{b}{4}\cdot\frac{1}{1+1}\cdot 1^2\end{align}$$

$\endgroup$
2
$\begingroup$

Hint: Use trigonometry identities:

$$\sin(x)=\cos\left(\frac{\pi}{2}-x\right)\\\sin\frac{t}{2}=\sqrt{\frac{1}{2}(1-\cos t)}$$

Specifically, set $t=\frac{\pi}{2}-x$ then you want:

$$\lim_{t\to 0} \frac{b(1-\cos t)}{4t^2}$$

Then use the trig identities above, replacing $1-\cos t$.

$\endgroup$
1
$\begingroup$

If $b = 0$, then there is nothing to work out; let $b \neq 0$. But $$ \lim_{x \to \pi/2} \frac{b(1 - \sin x)}{(\pi - 2x)^{2}} = b\lim_{h \to 0}\frac{1 - \sin (h + \pi/2)}{4h^{2}} = b\lim_{h \to 0}\frac{1 - \sin h \cos (\pi /2) - \cos h \sin (\pi/2)}{4h^{2}}\\ = b\lim_{h \to 0}\frac{1 - \cos h}{4h^{2}} = b\lim_{h \to 0}\frac{\frac{h^{2}}{2} + o(h^{2})}{4h^{2}} = \frac{b}{8}. $$

$\endgroup$
1
$\begingroup$

we get $$\frac{(1-\sin(x))(1+\sin(x))}{(1+\sin(x))(\pi-2x)^2}$$ with $t=\pi-2x$ we get $$\frac{(\sin(t/2))^2}{4(\frac{t}{2})^2}$$

$\endgroup$
0
$\begingroup$

I've always hated limits that don't approach zero, so my first step is to make the limit approach zero by expressing everything in terms of $x-\frac{\pi}{2}$, then make the subsitution $\theta = x - \frac{\pi}{2}$.

$$\begin{array}{lll} \displaystyle\lim_{x\to\pi/2}\frac{b(1-\sin x)}{(\pi-2x)^2}&=&\displaystyle\lim_{x\to\pi/2}\frac{b(1-\sin ((x-\frac{\pi}{2})+\frac{\pi}{2}))}{(\pi-2((x-\frac{\pi}{2})+\frac{\pi}{2}))^2}\\ &=&\displaystyle\lim_{\theta\to 0}\frac{b(1-\sin (\theta+\frac{\pi}{2}))}{(\pi-2(\theta+\frac{\pi}{2}))^2}\\ &=&\displaystyle\lim_{\theta\to 0}\frac{b(1-\sin (\theta+\frac{\pi}{2}))}{4\theta^2}\\ \end{array}$$

Can you take it from here?

Hint: $\sin (A+B)=\sin A \cos B + \cos A \ sin B$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.