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I would like to evaluate the following limit: $$\lim_{x \to 0}\frac{\tan^2x-x^2}{x^2\tan^2x}$$

I tried breaking the limit as:

$$\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$$

Writing tan is terms of sin and cos, this becomes:

$$\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{\cos^2x}{\sin^2x×x^2/x^2}\right)$$

Using standard limit of $\sin x/x$, we obtain:

$$\lim_{x \to 0}\frac{1-\cos^2x}{x^2}$$

Since $1-\cos^2x=\sin^2x$, using standard limits again we get 1 as the answer. But the correct answer is $2/3$. Please explain what is going wrong.

Kenny Wong
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Vedaansh Agarwal
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    Limit algebra (eg: the breaking step) is only valid when the broken limits exist by themself. Also this quesiton has been asked more than million times on this site. – tryst with freedom May 10 '23 at 09:30
  • Understood, thank you! – Vedaansh Agarwal May 10 '23 at 09:33
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    To be specific, the problem is not in the first step (“breaking”), since the limit there is still taken on the whole expression. It's rather with the step where the subexpression $\sin(x)/x \to 1$. You can't let some $x$ tend to zero while keeping other $x$ untouched, since all occurrences of $x$ stand for the same number! – Hans Lundmark May 10 '23 at 09:52
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    The mistake is when replacing a subexpression with its limit: $$\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{\cos^2x}{x^2\left(\frac{\sin x}x\right)^2}\right)\ne\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{\cos^2x}{x^2\left(1\right)^2}\right).$$ Such a mistake is more visible in the following: $$\lim_{x\to0}\frac xx\ne\lim_{x\to0}\frac0x$$ ($1\ne0$). – Anne Bauval May 10 '23 at 09:54
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    Hint: use difference of squares to express in terms of $\tan x/x$ and $(\tan x - x)/x^3$, whose limits both exist as $x\rightarrow 0$. – eyeballfrog May 10 '23 at 12:55
  • Thanks, makes a lot more sense now – Vedaansh Agarwal May 10 '23 at 16:04
  • Good point @HansLundmark – tryst with freedom May 10 '23 at 17:47

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