# Getting 2 answers for a limit

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

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 $$x\to0$$.
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 $$\displaystyle 2\cdot {-1\over 6}=-{1\over 3}.$$
If you observer the last step carefully, you'll notice that $$\frac {1}{x^2}$$ diverges as $$x$$ approaches 0. Therefore, $$\lim_{x\to 0}\frac{1}{x^2} - \lim_{x\to 0}\frac{1}{x^2} \neq 0$$