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?


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 $1/\sin(bx^2)<-1$.

  • $\begingroup$ wait... can you explain how that works, like you basically multiplied it by bx^2/ax^2, and it is still undefined when you sub in 0, thanks $\endgroup$ – Spongebob Squareroot-pants Oct 5 '15 at 15:39
  • $\begingroup$ @SpongebobSquareroot-pants Yes, but you surely know that $\lim\limits_{x\to 0}\dfrac{\sin x}{x}=1$. $\endgroup$ – egreg Oct 5 '15 at 15:40
  • $\begingroup$ @SpongebobSquareroot-pants When taking a limit, the point is not implicated. Therefore, there was no division by $0$. $\endgroup$ – Mark Viola Oct 5 '15 at 15:40
  • $\begingroup$ isn't that basically still Hospital? How do you show $\frac{sin\left(x\right)}{x}\stackrel{x\rightarrow 0}{\rightarrow} 1$? $\endgroup$ – Max Oct 5 '15 at 15:45
  • $\begingroup$ @Max That's a basic limit, used for finding the derivative of sine and cosine. $\endgroup$ – egreg Oct 5 '15 at 15:48

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

$$\lim_{x\to 0}\frac{ax^2\cos (ax^2)}{bx^2}=\frac ab$$


$$\lim_{x\to 0}\frac{ax^2}{bx^2\cos(bx^2)}=\frac ab$$

Thus, by the squeeze theorem,

$$\lim_{x\to 0}\frac{\sin(ax^2)}{\sin(bx^2)}=\frac ab$$

  • $\begingroup$ Thanks for providing such a nice alternate solution. +1 $\endgroup$ – Shailesh Oct 5 '15 at 16:07
  • $\begingroup$ @Shailesh You're welcome! My pleasure. $\endgroup$ – Mark Viola Oct 5 '15 at 16:11

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