# Thread: limits tan^2x and cot^x

1. ## limits tan^2x and cot^x

I split it up into the limit as x --> pi of

tan^2x + cot^2x

tan^2x = 1 - 1/cos^2x

in the limit x --> pi 1--1 = 2

for the cot^2 one

cot^2x = 1/tan^2x

i used the double angle formula and simplified to

$\frac{1}{1-\frac{2tanx}{tan2x}}$

then the infinities of tanx/tan2x cancel out so your left with 1/(1-2) = -1

i don't know how to more formally do the cot one!

2. ## Re: limits tan^2x and cot^x

[QUOTE=Applestrudle;817376]

I split it up into the limit as x --> pi of

tan^2x + cot^2x[quote]
??? $\frac{cos^2(x)}{sin^2(x)}= cot^2(x)$, not " $tan^2(x)$" and $\frac{cos^3(x)}{sin^2(x)}= cos(x)cot^2(x)$, not $cot^2(x)$. Did you not notice that, at $x=\frac{\pi}{4}$ the first is $1+ \frac{\sqrt{2}}{2}$ but you "split" formula has value 2?
I think you would be better advised to write it as $cos^2(x)\left(\frac{1+ cos(x)}{sin^2(x)}\right)= cos^2(x)\left(\frac{1+ cos(x)}{1- cos^2(x)}\right)$.

tan^2x = 1 - 1/cos^2x

in the limit x --> pi 1--1 = 2

for the cot^2 one

cot^2x = 1/tan^2x

i used the double angle formula and simplified to

$\frac{1}{1-\frac{2tanx}{tan2x}}$

then the infinities of tanx/tan2x cancel out so your left with 1/(1-2) = -1

i don't know how to more formally do the cot one!

3. ## Re: limits tan^2x and cot^x

[QUOTE=HallsofIvy;817379][QUOTE=Applestrudle;817376]

I split it up into the limit as x --> pi of

tan^2x + cot^2x
??? $\frac{cos^2(x)}{sin^2(x)}= cot^2(x)$, not " $tan^2(x)$" and $\frac{cos^3(x)}{sin^2(x)}= cos(x)cot^2(x)$, not $cot^2(x)$. Did you not notice that, at $x=\frac{\pi}{4}$ the first is $1+ \frac{\sqrt{2}}{2}$ but you "split" formula has value 2?
I think you would be better advised to write it as $cos^2(x)\left(\frac{1+ cos(x)}{sin^2(x)}\right)= cos^2(x)\left(\frac{1+ cos(x)}{1- cos^2(x)}\right)$.
okay a lot of stupid mistakes,

I get cot^2x goes to zero and cosxcot^2x goes to zero too