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Find $L =\lim\limits_{x\to0}\left(\dfrac{1}{x^2}-\cot^2{x}\right).$

The answer is $\dfrac23$, which I was able to obtain using L'Hopital. However, I originally took the following steps and couldn't find what is wrong. Please advise. Thank you!

As $x\to0$, $\sin{x} \sim x$ and $\cos{x} \sim 1-\dfrac{x^2}{2}$. Thus, $$L=\lim\limits_{x\to0}\left(\dfrac{1}{x^2}-\dfrac{(1-\frac{x^2}{2})^2}{x^2}\right)=\lim\limits_{x\to0}\dfrac{1-(1-x^2+\frac{x^4}{4})}{x^2} = \lim\limits_{x\to0}\dfrac{x^2-\frac{x^4}{4}}{x^2}=1.$$

Bowei Tang
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ten_to_tenth
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    $\sin x\sim x$ is insufficient (try to add the $x^3$ term...). – metamorphy Nov 12 '24 at 09:21
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    Your argument don't work because:

    If $;f_1\sim g_1;$ and $;f_2\sim g_2\quad(x\to a),;$ this does NOT implies $;f_1\pm f_2\sim g_1 \pm g_2\quad(x\to a)$.

     – gpassante Nov 15 '24 at 07:58

3 Answers3

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This is something that you have done in the problem: $$\lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^4}=\lim_{x\to 0}\frac{x^2-x^2(1-\frac{x^2}{2})^2}{x^4}=\lim_{x\to 0}\frac{x^4-\frac{x^6}{4}}{x^4} =1.$$ However, notice that $\sin x=x-\frac{x^3}{3!}+o(x^3)$ and thus $$\sin^2(x)=\left(x-\frac{x^3}{3!}+o(x^3)\right)^2=x^2-\frac{x^4}{3}+o(x^4).$$

So, if you replace $\sin x$ with $x$, you'll miss $-\frac{x^4}{3}$, but $\lim_{x\to 0}\frac{-\frac{x^4}{3}}{x^4}=-\frac{1}{3}$ exists.

Thus, the limit should be $1-\frac{1}{3}=\frac{2}{3}$.

Edit: I recommend you to combine the limit $\frac{1}{x^2}-\cot^2x$ to get $\frac{\sin^2x-x^2\cdot \cos^2x}{x^2\sin^2x}$ first, then for the denominator, you can use $\sin x\sim x$ to get $\frac{\sin^2x-x^2\cdot \cos^2x}{x^4}$. Then, you should be careful of what to do next. You should consider the expansion up to $x^4$ since for $x^n(n\geq 5)$, the limit will be $0$. For $\sin x$, I've shown the reason above; For $\cos^2x$, $\cos x\sim 1-\frac{x^2}{2}$ is enough since if you add $\frac{x^4}{2}$ then square it, you'll get some new terms with at least $x^4$ compared with $(1-\frac{x^2}{2})^2$. And with a multiple of $x^2$, it will generate degrees of at least $6$, so we don't need to consider more.

Edit: For $x^2\sin^2x$, if you try to expand, you'll get $x^4-\frac{x^6}{3}+o(x^6)$. In layman's terms, $x^4$ dominates. I'll give a simple example: $\lim_{x\to 0}\frac{3x^3-x^7}{x^3-x^4}=3$ since it is the same as $\lim_{x\to 0}\frac{3-x^4}{1-x}$. Thus, $x^4$ in the denominator doesn't really do anything. In conclusion, if $x\to 0$, then you only need to consider the lowest degree as $$\frac{...}{x^4}\frac{x^4}{x^4-\frac{x^6}{3}+o(x^6)}=\frac{..}{x^4}\ \text{as}\ x\ \text{approaches}\ 0.$$

Bowei Tang
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  • Thanks a lot! Could you please give me some tips on determining at which degree to stop? I redid the problem using the Maclaurin expansions of $\sin{x}$ and $\cos{x}$ as you suggested and then identified the expansions for $\sin^2{x}$ and $\cos^2{x}$ accordingly, but I don't know why I should not go beyond $x^4$. Thank you! – ten_to_tenth Nov 12 '24 at 10:32
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    @ten_to_tenth Please see the edit. – Bowei Tang Nov 12 '24 at 10:45
  • Thanks! Tbh I'm a little confused. Why are we allowed to "fix" $\sin{x}$ as $x$ in the denominator but have to consider $\sin{x}$ in the numerator more carefully? – ten_to_tenth Nov 12 '24 at 12:06
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    @ten_to_tenth Updated. Hope this helps! – Bowei Tang Nov 12 '24 at 12:45
  • Thank you very much! It does! – ten_to_tenth Nov 12 '24 at 18:02
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Hint:

$$\dfrac1{x^2}-\dfrac1{\tan^2x}=\dfrac{\tan x-x}{x^3}\cdot\dfrac{\tan x+x}x\cdot\left(\dfrac x{\tan x}\right)^2$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion

lab bhattacharjee
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With the change of variable $x=\arctan(t)$, continous at $0$,

$$\lim_{t\to0}\left(\frac1{\arctan^2(t)}-\dfrac1{t^2}\right)=\lim_{t\to0}\left(\frac1{\left(t-\dfrac{t^3}3+\cdots\right)^2}-\dfrac1{t^2}\right)=\lim_{t\to0}\left(\dfrac{1+\dfrac{2t^2}3+\cdots}{t^2}-\dfrac1{t^2}\right)=\frac23.$$