1. ## Trig Limit

Is there a faster way to solve this limit, $\displaystyle\lim_{x\to{0}}\frac{1}{x^2}-cot^2\;x$ other than using L'Hopital's Rule twice?

2. You can always fall back on the Taylor series

The first few terms is about all you need.

$cot^{2}(x)=\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)$

Now, you have:

$\frac{1}{x^{2}}-\left(\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)\right)$

$\lim_{x\rightarrow{0}}\frac{2}{2x^{2}+3}=\frac{2}{ 3}$

3. Originally Posted by galactus
You can always fall back on the Taylor series

The first few terms is about all you need.

$cot^{2}(x)=\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)$

Now, you have:

$\frac{1}{x^{2}}-\left(\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)\right)$

$\lim_{x\rightarrow{0}}\frac{2}{2x^{2}+3}=\frac{2}{ 3}$
Could you explain to me the idea of Taylor's series - what is it, how to use it....

4. No, I am sorry. Too extensive a topic. Look it up anywhere. Books, Google, etc. If you're not familiar, then I suppose another method is called for.

5. Originally Posted by polymerase
Is there a faster way to solve this limit, $\displaystyle\lim_{x\to{0}}\frac{1}{x^2}-cot^2\;x$ other than using L'Hopital's Rule twice?
Hmm, not really. I have always thought L'Hopital's trick was very fast.

6. Originally Posted by colby2152
Hmm, not really. I have always thought L'Hopital's trick was very fast.
there are times when galactus' solution is a lot faster than l'Hopital's.

not sure which is the case here, never tried doing the problem. it may take a while to come up with the Taylor series for $\cot^2 x$ though

7. I agree with Jhevon, and it is also more interesting if you try to avoid using L' Hôpital.

A bit about Taylor polynomials: the nth degree polynomial of $f(x)$, $T_n$, centered in , let's say "a", has the following property: $T_n(a)=f(a)$ , $T_n'(a)=f'(a)$ ... $T_n^{(n)}(a)=f^{(n)}(a)$.
It turns out to be a good approximation of the function (in a certain interval)

The remainder is given by $f(x)-T_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}\cdot{(x-a)^{n+1}}$ for some $c\in(a, x)$

You can check that: $T_n=f(a)+\sum_{k=1}^n{\left(\frac{f^{(k)}(a)}{k!}\ cdot{(x-a)^k}\right)}$