1
$\begingroup$

How to compute

$$\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=-\dfrac{3}{2}$$

I'm interested in more ways of computing limit for this expression.

My Thoughts

at least i tried to use L'Hospital's rule but with no luck

\begin{align} &\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\lim _{x\to \:0}\dfrac{2x\left(\dfrac{\cos \left(x^2\right)}{\sin \left(x^2\right)+1}-1\right)}{\dfrac{2\arcsin \left(x\right)}{\sqrt{1-x^2}}-2x}\\ &=\lim _{x\to \:0}\dfrac{2\left(\dfrac{\cos \left(x^2\right)\left(\sin \left(x^2\right)+1\right)-2x^2\left(\cos ^2\left(x^2\right)+\sin ^2\left(x^2\right)+\sin \left(x^2\right)\right)}{\left(\sin \left(x^2\right)+1\right)^2}-1\right)}{2\left(\dfrac{x\arcsin \left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{1}{1-x^2}-1\right)} \end{align} Note this limit was taken from competition mathematics so i can't go with L'Hopital because each time i use it i got big terms that i have to derivative for the next time

$\endgroup$
2
$\begingroup$

We will use the following results $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\text{ (via LHR)}$$ and $$\lim_{x \to 0}\frac{\arcsin x}{x} = \lim_{t \to 0}\frac{t}{\sin t} = 1\text{ (by putting }t = \arcsin x)$$ We can then proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - x^{2}}{(\arcsin x)^{2} - x^{2}}\notag\\ &= \lim_{x \to 0}\frac{\log(1 + \sin(x^{2})) - x^{2}}{x^{4}}\cdot\frac{x^{4}}{(\arcsin x)^{2} - x^{2}}\notag\\ &= \lim_{z \to 0}\frac{\log(1 + \sin z) - z}{z^{2}}\cdot\lim_{x \to 0}\frac{x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\text{ (putting }z = x^{2})\notag\\ &= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z + \sin z - z}{z^{2}}\notag\\ &\,\,\,\,\times\lim_{x \to 0}\frac{x}{\arcsin x}\cdot\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\notag\\ &= \lim_{z \to 0}\left(\frac{\log(1 + \sin z) - \sin z}{z^{2}} - z\cdot\frac{z - \sin z}{z^{3}}\right)\notag\\ &\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{\arcsin x - x}\notag\\ &= \lim_{z \to 0}\left(\frac{\log(1 + \sin z) - \sin z}{z^{2}} - 0\cdot\frac{1}{6}\right)\notag\\ &\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{x^{3}}{(\arcsin x)^{3}}\cdot\frac{(\arcsin x)^{3}}{\arcsin x - x}\notag\\ &= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z}{\sin^{2}z}\cdot\frac{\sin^{2}z}{z^{2}}\notag\\ &\,\,\,\,\times\lim_{x \to 0}\frac{\arcsin x}{\arcsin x + x}\cdot\frac{(\arcsin x)^{3}}{\arcsin x - x}\notag\\ &= \lim_{z \to 0}\frac{\log(1 + \sin z) - \sin z}{\sin^{2}z}\cdot\lim_{t \to 0}\frac{t}{t + \sin t}\cdot\frac{t^{3}}{t - \sin t}\text{ (putting }t = \arcsin x)\notag\\ &= 6\lim_{u \to 0}\frac{\log(1 + u) - u}{u^{2}}\cdot\lim_{t \to 0}\dfrac{1}{1 + \dfrac{\sin t}{t}}\text{ (putting }u = \sin z)\notag\\ &= 3\lim_{u \to 0}\frac{\log(1 + u) - u}{u^{2}}\notag\\ &= 3\lim_{u \to 0}\dfrac{\dfrac{1}{1 + u} - 1}{2u}\text{ (via LHR)}\notag\\ &= -\frac{3}{2}\lim_{u \to 0}\frac{1}{1 + u}\notag\\ &= -\frac{3}{2}\notag \end{align} This involves 2 applications of LHR. The use of LHR can be replaced with Taylor series also. Note that both LHR and Taylor series are powerful tools to evaluate limits but they should always be used along with algebraic simplification of limit expression (and use of standard limits) otherwise their usage is a bit complicated.

$\endgroup$
  • $\begingroup$ You are right use LHR with algebraic simplification of limit expression and standard limits makes things easy $\endgroup$ – Educ Sep 24 '15 at 6:50
  • 1
    $\begingroup$ @Educ: However its rather unfortunate that students jump to LHR when they see $0/0$. They first need to simplify the expression using limit theorems and standard limits. $\endgroup$ – Paramanand Singh Sep 24 '15 at 7:28
  • $\begingroup$ especially in math-contest $\endgroup$ – Educ Sep 24 '15 at 7:29
  • 1
    $\begingroup$ @Educ: If such questions are asked in contest (provided you don't have to show the working) then also the above method is fast and the answer is almost instantaneous. If you need to show the working then Taylor is much faster as you need to write very less. $\endgroup$ – Paramanand Singh Sep 24 '15 at 7:34
  • $\begingroup$ Thank you, yes of cours in that kind of contest MCQ they need just your final answer not showing proof so we must search for faster way to answer them but i just would like to see all proofs to understand much more $\endgroup$ – Educ Sep 24 '15 at 7:42
2
$\begingroup$

You must use power series development at order $4$:

  • $\sin x= x+o(x^2)$, hence $\sin x^2=x^2+o(x^4)$

  • $\ln(1+u)=u-\dfrac{u^2}2+o(u^2) $, hence $$\ln(1+\sin x^2)=\ln\bigl(1+ x^2+o(x^4)\bigr)=x^2-\frac{x^4}2+o(x^4)$$

  • $\arcsin x=x+\dfrac12\dfrac{x^3}3+o(x^4)$, hence $(\arcsin x)^2=x^2+\dfrac{x^4}3+o(x^4)$

Grouping all the results we get: $$\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\frac{-\dfrac{x^4}2+o(x^4)}{\dfrac{x^4}3+o(x^4)}=-\frac32+o(1)\to-\frac32.$$

$\endgroup$
  • $\begingroup$ that's Clean and Fast $\endgroup$ – Educ Sep 23 '15 at 21:22
0
$\begingroup$

\begin{align} \lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2} &= \lim _{x\to \:0}\frac{\sin \left(x^2\right)-\frac{\sin^2(x^2)}{2} + O(x^6)-x^2}{\left(\arcsin \:x\right)^2-x^2} \end{align}

(The above is true as $\ln(1+x) \approx x-x^2/2 +O(x^3)$ when $x$ is small)

\begin{align} \lim _{x\to \:0}\frac{\sin \left(x^2\right)-\frac{\sin^2(x^2)}{2} + O(x^6)-x^2}{\left(\arcsin \:x\right)^2-x^2} &= \lim _{x\to \:0}\frac{x^2 -x^4/2 + O(x^{6})-x^2}{\left(x + \frac{x^3}{6}+O(x^5)\right)^2-x^2} \\ &=\lim _{x\to \:0}\frac{\frac{-x^4}{2}}{\frac{2x^4}{6}}\\ &= -\frac{3}{2} \end{align}

(I used Taylor expansions)

$\endgroup$
  • 2
    $\begingroup$ Unfortunately the answer has errors. For example, $\sin x^2\ne \sin^2 x$. $\sin x^2=x^2-\frac16x^6+O(x^{10})$. $\sin ^2x=(x-\frac16 x^3+O(x^5))^2=x^2-\frac13x^4+O(x^6)$. And please explain the justification to arrive at the second to last equality. $\endgroup$ – Mark Viola Sep 23 '15 at 20:59
  • $\begingroup$ the answer is $-\dfrac{3}{2}$ so please would you provide more details $\endgroup$ – Educ Sep 23 '15 at 21:09
  • $\begingroup$ Thanks. I made the changes and edited the answer. Hope it helps! $\endgroup$ – chandu1729 Sep 23 '15 at 21:15

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.