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How to calculate limit of 1/x^2 - cot^2 x when x approaches 0. When I use cos^x and sin^x for the above function i can't get a proper answer. When I use l'hospital rule I always get 1/0 type answer. Could you please help me to solve this. is it correct to get an answer like 1/0? Can you please give me the correct answer

3 Answers 3

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Using Taylor's polynomial at order $4$ will do:

$$\tan x=x+\frac{x^3}3+o(x^4),\enspace\text{hence}\quad \tan^2x=\Bigl(x+\frac{x^3}3+o(x^4)\Bigr)^2=x^2+\frac{2x^4}3+o(x^4),$$ so that $$\frac1{\tan^2x}=\frac1{\phantom{t^2}x^2}\cdot\frac1{1+\dfrac{2x^2}3+o(x^2)}=\frac1{x^2}\Bigl(1-\dfrac{2x^2}3+o(x^2)\Bigr)=\frac1{x^2}-\dfrac23+o(1),$$ so that finally $$\frac1{x^2}-\cot^2x=\frac23+o(1).$$

Added: a (less simple) solution using L'Hospital's rule.

Rewrite the expression as $$\frac 1{x^2}-\frac 1{\tan^2x}= \frac{\tan^2x-x^2}{x^2\tan^2x}=\frac{\tan x-x}{x^2\tan x}\cdot\frac{\tan x+x}{\tan x}=\frac{\tan x-x}{x^2\tan x}\Bigl(1+\frac x{\tan x}\Bigr).$$ The second factor tends to $1+1=2$. Let's take care of the first factor: by L'Hospital's rule, $$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}=\lim_{x\to 0}\frac{\tan^2 x}{2x\tan x+x^2(1+\tan^2x)}=\lim_{x\to 0}\frac{1}{\dfrac{2x}{\tan x}+\dfrac{x^2}{\tan^2x}(1+\tan^2x)}=\frac 1{2+1}.$$ As a conclusion, the limit is $\dfrac 13\cdot 2$.

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    +1, but i think OP wanted to apply L'Hospital's Rule... – 2017-01-30
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    I didn't understand the answer. In our college we didn't study something like Taylor's polynomial. Can't we solve this using only l'hospital 's rule? – 2017-01-30
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    L'Hospital's rule should be avoided as much as possible. Students generally forget to check its conditions of validity, so that it may lead to false answers, and anyway it is equivalent to Taylor's polynomial at order $1$. – 2017-01-30
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    I didn't get a proper answer yet. Please help me to find the final answer of this. – 2017-01-30
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    Thanks alot for your answer – 2017-01-31
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HINT $$ \lim_{x \to 0} \frac{1}{x^2} - \frac{\cos^2 x}{\sin^2 x} = \lim_{x \to 0} \frac{\sin^2 x - x^2 \cos^2 x}{x^2 \sin^2 x}, $$ which is of the form 0/0 suitable for application of L'Hospital's Rule...

HINT 2 To apply L'Hospital's Rule, remember that $$ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} $$ under nice conditions, which hold in your case. So differentiate both numerator and denominator and take the limit of their ratio, and you will be good to go.

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    I can't get an answer from this equation. I can't understand how to solve – 2017-01-30
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    @user6333477 please see update – 2017-01-30
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    In this case also I get 1/0. Then is the answer infinity? – 2017-01-30
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    @user6333477 if you did the arithmetic correctly, yes – 2017-01-30
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    Please can you give me the answer which you got. The way of getting the answer. Please. – 2017-01-30
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    @user6333477 I am trying hard not to do this for you. Please post your work and I'll be happy to guide you further. – 2017-01-30
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Note that, when $x\to 0$, $x^2\to 0$ but $\cot^2 x \to +\infty$. So, if you write the limit $$\lim_{x\to 0} \left(\frac{1}{x^2}-\cot^2 x\right)=\lim_{x\to 0} \frac{1-x^2\cot^2 x}{x^2}$$ you will create an indeterminate form $0\cdot\infty$. So, you can't substitute and get $\frac10$ in this case.

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    Can you please give me an answer for this function. I can't get a good answer – 2017-01-30
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    I assume you know how to apply l'Hospital's rule? – 2017-01-30
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    Yes I know. But I can't get the final answer. I don't know how to solve this. Can you give me the answer and the way of solving this – 2017-01-30