Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$

Question

$\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$

I wanted to use L' Hopital's rule so I wrote the term as: $$\frac{\sin^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)}{\cos^2{x}}$$ and found the first derivative of the denumaerator & denominator: $\frac{{\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\sin{x}\cos{x}}{2\cos{x}(-\sin{x})}$

$$\lim_{x\to \frac{\pi}{2}}{\left(-2\left({\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)\sec{x}\csc{x}\right)}$$

But I didn't know how to continue. I drew a graph in Geogebra:

Answer 1

To simplify we have that

$$\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)=$$

$$\sin^2{x}\frac{\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}{\cos^2 x}$$

and since $t=\sin^2 x \to 1$ we can consider the limit

$$\lim_{t\to 1} \frac{\sqrt{2t^2+3t+4}-\sqrt{t^2+6t+2}}{1-t^2}$$

which can be easily solved by rationalization or also by L'Hopital if you prefer to obtain

$$\lim_{t\to 1} \frac{\sqrt{2t^2+3t+4}-\sqrt{t^2+6t+2}}{1-t^2}=\lim_{t\to 1} \frac{\frac{4t+3}{2\sqrt{2t^2+3t+4}}-\frac{2t+6}{2\sqrt{t^2+6t+2}}}{-2t}=\frac1{12}$$

Answer 2

Hint:

Set $\sin x=s$

Rationalize the numerator

$$s^2\cdot\dfrac{\sqrt{2s^2+3s+4}-\sqrt{s^2+6s+2}}{1-s^2}$$

$$=s^2\cdot\dfrac{(s^2-3s+2)}{1-s^2}\cdot\dfrac1{\cdots}$$

As $s\to1,s\ne1$ so cancel out $s-1$ and set $s\to1$

Answer 3

Instead of using L'Hospital, multiply the top and bottom by conjugate: $$\frac{\sin^2{x}\Big((2\sin^2{x}+3\sin{x}+4)-(\sin^2{x}+6\sin{x}+2)\Big)}{\cos^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\Big)}=\\ \require{cancel}\frac{\sin^2{x}(\cancel{1-\sin x})(2-\sin x)}{(\cancel{1-\sin x})(1+\sin x)\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\Big)}.$$ And now plug $x=\frac{\pi}{2}$ to get: $$\frac{1^2\cdot 1}{(1+1)(3+3)}=\frac1{12}.$$