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The limit in question is $$\lim_{x\to 0} \frac{\sin^2(x) - x^2}{x^2 \sin^2(x)} $$

If we solve it via L'Hopitals method we get the answer to be $\frac{-1}{3}$

While we could also simplify it $$\lim_{x\to 0} \frac{\sin^2(x) - x^2}{x^2 \sin^2(x)} = \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0}\frac{1}{x^2}\frac{x^2}{\sin^2(x)} $$

$$\lim_{x\to 0} \frac{\sin^2(x) - x^2}{x^2 \sin^2(x)} = \lim_{x\to 0}\frac{1}{x^2} - \lim_{x\to 0}\frac{1}{x^2}.\lim_{x\to 0}\frac{x^2}{\sin^2(x)} $$

$$\lim_{x\to 0} \frac{\sin^2(x) - x^2}{x^2 \sin^2(x)} = \lim_{x\to 0}\frac{1}{x^2} - \lim_{x\to 0}\frac{1}{x^2} = 0$$

Which is different from the above answer

Wolfram Alpha also says the answer is $\frac{-1}{3}$

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    $\begingroup$ sum of limits is not the same as limit of sums if limits do not exist. $\endgroup$
    – Gregory
    Commented Sep 18, 2023 at 13:48
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    $\begingroup$ You can't seperate them comrade $\endgroup$ Commented Sep 18, 2023 at 14:00
  • $\begingroup$ Eleanora, are you interested in another solution without using L'Hopital’s method ? $\endgroup$
    – Angelo
    Commented Sep 18, 2023 at 14:53
  • $\begingroup$ Why don't you apply $a^2-b^2=(a+b)(a-b)$ ? in the numerator ? $\endgroup$ Commented Sep 18, 2023 at 15:08
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    $\begingroup$ A limit of $\infty-\infty$ is itself an indeterminate form and isn't necessarily 0. $\endgroup$ Commented Sep 18, 2023 at 15:15

3 Answers 3

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By Taylor's formula, $$\sin^2(x)=x^2-\frac{x^4}{3}+O(x^6)$$ as $x\to0$. This is enough to see that, $$\frac{\sin^2(x) - x^2}{x^2 \sin^2(x)}\sim-\frac{1}{3}$$ as $x\to0$. As the comments say, you deal with limits that do not exist when you separate them. For example $1/x^2$ diverges as $x\to0$.

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We have $${\sin^2x-x^2\over x^2\sin^2x}={\sin x+x\over x}\,{\sin x-x\over x^3}\,{x^2\over \sin^2x}$$ The first factor tends to $2,$ while the last one to $1.$ The factor in the middle can be calculated by applying the l'Hospital rule once $$\lim_{x\to 0}{\sin x-x\over x^3}\overset{\rm (H)}{=}\lim_{x\to 0}{\cos x-1\over 3x^2}= \lim_{x\to 0}{-\sin^2x\over 3x^2}{1\over 1+\cos x}=-{1\over 6}$$ So the result is equal $\displaystyle 2\cdot {-1\over 6}=-{1\over 3}.$

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If you observer the last step carefully, you'll notice that $\frac {1}{x^2}$ diverges as $x$ approaches 0. Therefore, $$\lim_{x\to 0}\frac{1}{x^2} - \lim_{x\to 0}\frac{1}{x^2} \neq 0$$

What you can do to solve this question, as others have suggested, is use the Taylor series or the L' Hospital rule.

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