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I know that some of the $\infty - \infty$ limits can be solved without using l'Hopital's rule. In such cases, you would usually be able to either rationalize (or derationalize if that is what its called) in the case of rational power expressions or you would take the lcm and cancel the common factors or simply apply squeeze theorem or Taylor's theorem. But I came across this question: $$ \lim_{x\to 0} \left(\,\frac{1}{x^{2}} - \frac{1}{(\sin x)^{2}}\right) $$

I am able to solve this using l'Hôpital's rule, and I get the limit at $0$ as $\frac{-1}{3}$ but I am not able to solve it using any other method. So, is there any alternative way to solve this limit?

PS- I changed $\frac{1}{3}$ to $\frac{-1}{3}$ as the latter is the correct solution. Thnx for everyone who edited my question to make it better and thnx for pointing out my mistakes.

Spime
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  • Welcome to Mathematics Stack Exchange. Did you mean $-\frac13$? – J. W. Tanner Nov 26 '23 at 19:18
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    There should be a new stackexchange page - life without L'Hospital's rule. – Dietrich Burde Nov 26 '23 at 19:21
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    $\dfrac{\sin^2x-x^2}{x^2\sin^2x}=\dfrac{(x-\frac16x^3\cdots)^2-x^2}{x^2\sin^2x}=\dfrac{(x^2-\frac13x^4\cdots)-x^2}{x^2\sin^2x}\approx\dfrac{-\frac13x^4}{x^4}$ – J. W. Tanner Nov 26 '23 at 19:27
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    @DietrichBurde We do have [tag:limits-without-lhopital] – MJD Nov 26 '23 at 19:39
  • Cf. this (closed) question – J. W. Tanner Nov 26 '23 at 19:40
  • There are arguments using the power series for $\sin x,$ but those can be seen as implicit usages of L'Hopital (or, depending on how you look at it, L'Hopital is a way of encoding arguments involving power series.) – Thomas Andrews Nov 26 '23 at 19:49
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    (1) Did you know that LH rule was "Proved" without using LH rule ? (2) We can use the Exact Same Proof to get the limits , all without using LR rule ! (3) If we use LH rule to make a new theorem OY rule to get limits , then can we use OY rule here ? (4) What I am getting at is : When-ever we use LH rule to get a limit , we can always avoid LH rule to get the limit , by going back to Basics ! – Prem Nov 26 '23 at 19:58
  • Another way to calculate OP’s limit:$$\dfrac1{x^2}-\dfrac1{\sin^2!x}=-\dfrac x{\sin x}\left(\dfrac{x-\sin x}{x^3}\right)!\cdot!\left(\dfrac x{\sin x}+1\right)\overset{\text{as }x\to0}{\longrightarrow}$$ $$\overset{\text{as }x\to0}{\longrightarrow}-1!\cdot!\dfrac16!\cdot!\big(1+1\big)=-\dfrac13;.$$ – Angelo Nov 26 '23 at 20:19
  • @Prem can you link the proof here? Thnx for the info btw since I didn't know there was a proof like that. – Spime Nov 27 '23 at 13:25
  • @J.W.Tanner can you explain your solution a bit more? I didn't understand how you took care of the $(sinx)^{2}$ in the denominator. Also, how did you square an infinite series, I am still new to the taylor series and such and don't know if we can exponentiate such series. – Spime Nov 27 '23 at 13:29
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    Proof : https://www.macmillanlearning.com/studentresources/highschool/mathematics/rogawskiap2e/additionalproofs/proofoflhopitalsrule.pdf : https://math.stackexchange.com/questions/505535/proof-of-lhospitals-rule : https://proofwiki.org/wiki/L%27H%C3%B4pital%27s_Rule : You can use the Same Ideas to get the limit without using LH rule ! – Prem Nov 27 '23 at 13:33
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    The title doesn't match your question. – jjagmath Nov 27 '23 at 14:58
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    Also related: Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit? – user170231 Nov 27 '23 at 15:03

2 Answers2

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Writing the functions involved as Taylor polynomials will often help. It does in this case. It’s simpler if you use $\frac{1}{f} - \frac{1}{g} = \frac{g - f}{fg}$.

stange
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gnasher729
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$\lim\limits_{x\to0}\dfrac{\sin^2x-x^2}{x^2\sin^2x}=\lim\limits_{x\to0}\dfrac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}$

$=\lim\limits_{x\to0}\dfrac{(-\frac16x^3\cdots)(2x \cdots)}{x^4}\lim\limits_{x\to0}\dfrac x{\sin x}\lim\limits_{x\to0}\dfrac x{\sin x}$

$=-\frac13.$

J. W. Tanner
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