Solving limits without using L'Hôpital's rule

Question

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

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.

Answer 1

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 $1-\cos t$.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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 $\color{red}{\dfrac b8}.$

Answer 6

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