I had this question for math $$ \lim_{x\to 0}\frac{\sin{ax^2}}{\sin{bx^2}} $$
So I used squeeze theorem and got $-1<1/(\sin{bx^2})<1$ and I multiplied by $\sin{ax^2}$ and got $-\sin{ax^2}< \sin{ax^2}/\sin{bx^2}< \sin{ax^2}$
Which then equals $0 < \lim_{x\to 0}\sin{ax^2}/\sin{bx^2}<0$ and the answer I got is $0$, but the answer is supposed to be $a/b$
Where did I go wrong and how can you solve this without using l'Hôpital's rule?
On
Another way to "squeeze" the ratio is to use the inequalities
$$\left|x\cos x\right|\le\left|\sin x\right|\le \left|x\right|$$
Then, we have
$$\left|\frac{ax^2\cos (ax^2)}{bx^2}\right|\le\left|\frac{\sin(ax^2)}{\sin(bx^2)}\right|\le\left|\frac{ax^2}{bx^2\cos(bx^2)}\right|$$
In the limit as $x\to 0$, we have then that
$$\lim_{x\to 0}\frac{ax^2\cos (ax^2)}{bx^2}=\frac ab$$
and
$$\lim_{x\to 0}\frac{ax^2}{bx^2\cos(bx^2)}=\frac ab$$
Thus, by the squeeze theorem,
$$\lim_{x\to 0}\frac{\sin(ax^2)}{\sin(bx^2)}=\frac ab$$
Use $(\sin x)/x$: $$ \lim_{x\to0}\frac{\sin(ax^2)}{\sin(bx^2)}= \lim_{x\to0}\frac{\sin(ax^2)}{ax^2}\frac{bx^2}{\sin(bx^2)}\frac{a}{b} $$
Where did you go wrong? The inequality $-1<1/\sin(bx^2)<1$ is false for all $x$: if $0<\sin(bx^2)<1,$ then $1/\sin(bx^2)>1$; if $-1<\sin(bx^2)<0,$ then $1/\sin(bx^2)<-1$.