Limit of trig fcn: lim x-> pi / 4 [ tan(2x)] / [ cot(....

[peblez]

Find the value of lim x -- > pi / 4 [ tan ( 2 x )] / [ cot ( pi / 4 - x ) ]

i tried expanding and simplying but i just can't get the answer

 

[Deleted member 4993]

Re: Limit of a trig function

Find the value of lim x -- > pi / 4 [ tan ( 2 x )] / [ cot ( pi / 4 - x ) ]


This is inf/inf -> apply L'hospital - multiple times.


i tried expanding and simplying but i just can't get the answer

 

[peblez]

Re: Limit of a trig function

Yes, but i tried to do it that way but u'd have to do it a lot of times, soon you will be using the product rule 5 times or more, and gets very complicated.

 

[Deleted member 4993]

Re: Limit of a trig function

Yes, but i tried to do it that way but u'd have to do it a lot of times, soon you will be using the product rule 5 times or more, and gets very complicated.

Life is complicated.

However use trigonometric identities (eg 2 * sin(x) * cos(x) = sin(2x)) - and it will somewhat lessen the complication.

If you show what you tried - may be I can help to "uncomplicate".

 

[soroban]

Re: Limit of a trig function

Hello, peblez!

\(\displaystyle {\text{Find: }\;\lim_{x\to\frac{\pi}{4}} \frac{\tan(2x)}{\cot\left(\frac{\pi}{4} - x\right)}\)

$\displaystyle \text{Since: }\;\tan(2x) \:=\:\frac{2\tan x}{1-\tan^2\!x}$

. . $\displaystyle \text{and: }\;\cot\left(\frac{\pi}{4} - x\right) \:=\:\frac{1}{\tan\left(\frac{\pi}{4}-x\right)} \;=\;\frac{1 + \tan\frac{\pi}{4}\!\cdot\!\tan x}{\tan\frac{\pi}{4} - \tan x} \;=\;\frac{1 + \tan x}{1 - \tan x}$


$\displaystyle \text{then: }\;\frac{\tan(2x)}{\cot\left(\frac{\pi}{4}-x\right)} \;\;=\;\;\frac{\frac{2\tan x}{1-\tan^2\!x}}{\frac{1+\tan x}{1-\tan x}} \;\;=\;\;\frac{2\tan x}{(1-\tan x)(1+\tan x)}\cdot\frac{1-\tan x}{1+\tan x} \;\;=\;\;\frac{2\tan x}{(1+\tan x)^2}$


$\displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2} \;\;=\;\;\frac{2\cdot 1}{(1+1)^2} \;\;=\;\;\frac{2}{4}\;\;=\;\;\frac{1}{2}$