Test whether the following limit exists or not :

$$\lim_{x\to 0}\frac{\left[x^2\right]}{\sin x^2}$$

Here , $\displaystyle \lim_{x\to 0^-}\frac{\left[x^2\right]}{\sin x^2}=\lim_{x\to 0^-}\frac{0}{\sin x^2}=0$. And $\displaystyle \lim_{x\to 0^+}\frac{\left[x^2\right]}{\sin x^2}=\lim_{x\to 0^+}\frac{0}{\sin x^2}=0$. So limit exists and its value is $0$.

Is it correct ?

  • 1
    $\begingroup$ Yup that's right. $\endgroup$ – user223391 Mar 16 '16 at 5:52
  • 2
    $\begingroup$ The proof is correct because for $-1< x<1$ we have $[x^2]=0$, at least assuming that $[a]$ means the greatest integer $\le a$. (That is also written $\lfloor a\rfloor$.) $\endgroup$ – ForgotALot Mar 16 '16 at 6:01
  • $\begingroup$ As x approaches 0 from the negative direction isn't [x]=-1? $\endgroup$ – bulbasaur Mar 16 '16 at 6:25
  • $\begingroup$ but what about the $\frac{0}{0}$ form? what do you do about $\sin(x^2)$ $\endgroup$ – Airdish Mar 16 '16 at 6:26
  • 1
    $\begingroup$ @panja717 what does square braccet represents? $\endgroup$ – user5954246 Mar 16 '16 at 6:39

You cannot forget about the ${\sin x^2}$ in the denominator as this causes the fraction within the limit to evaluate to $\frac{0}{0}$, which is an indeterminate form. i.e. You cannot divide by $0$.

What we want to employ here is L'Hospital's Rule which, in layman's terms, allows us to take the derivative of both the numerator and denominator separately. We apply this rule anytime our fraction within a limit evaluates to a indeterminate form. Ex: $\frac{0}{0}$, $\frac{\infty}{\infty}$ , ...

In our case we can simplify the limit from: $$\lim_{x\to 0}\frac{x^2}{\sin x^2}$$ To: $$\lim_{x\to 0}\frac{2x}{2x\cos x^2}$$ The $2x$ in the numerator and denominator cancel to simplify the limit: $$\lim_{x\to 0}\frac{1}{\cos x^2}$$ Which evaluates to: $$\lim_{x\to 0}\frac{1}{\cos (0)^2} = \frac{1}{1} = 1$$ Therefore: $$\lim_{x\to 0}\frac{x^2}{\sin x^2} = 1$$

  • 3
    $\begingroup$ The function in the question is $[x^2]/\sin x^2$. $[x^2]=0$, when $|x|<1$, right? $\endgroup$ – Cm7F7Bb Mar 16 '16 at 7:47

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.