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Limit of trig fcn: lim x-> pi / 4 [ tan(2x)] / [ cot(....
- Thread starter peblez
- Start date
D
Deleted member 4993
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Re: Limit of a trig function
peblez said: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
D
Deleted member 4993
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Re: Limit of a trig function
peblez said: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".
Re: Limit of a trig function
Hello, peblez!
\(\displaystyle \text{Since: }\;\tan(2x) \:=\:\frac{2\tan x}{1-\tan^2\!x}\)
. . \(\displaystyle \text{and: }\;\cot\left(\frac{\pi}{4} - x\right) \:=\:\frac{1}{\tan\left(\frac{\pi}{4}-x\right)} \;=\;\frac{1 + \tan\frac{\pi}{4}\!\cdot\!\tan x}{\tan\frac{\pi}{4} - \tan x} \;=\;\frac{1 + \tan x}{1 - \tan x}\)
\(\displaystyle \text{then: }\;\frac{\tan(2x)}{\cot\left(\frac{\pi}{4}-x\right)} \;\;=\;\;\frac{\frac{2\tan x}{1-\tan^2\!x}}{\frac{1+\tan x}{1-\tan x}} \;\;=\;\;\frac{2\tan x}{(1-\tan x)(1+\tan x)}\cdot\frac{1-\tan x}{1+\tan x} \;\;=\;\;\frac{2\tan x}{(1+\tan x)^2}\)
\(\displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2} \;\;=\;\;\frac{2\cdot 1}{(1+1)^2} \;\;=\;\;\frac{2}{4}\;\;=\;\;\frac{1}{2}\)
Hello, peblez!
\(\displaystyle {\text{Find: }\;\lim_{x\to\frac{\pi}{4}} \frac{\tan(2x)}{\cot\left(\frac{\pi}{4} - x\right)}\)
\(\displaystyle \text{Since: }\;\tan(2x) \:=\:\frac{2\tan x}{1-\tan^2\!x}\)
. . \(\displaystyle \text{and: }\;\cot\left(\frac{\pi}{4} - x\right) \:=\:\frac{1}{\tan\left(\frac{\pi}{4}-x\right)} \;=\;\frac{1 + \tan\frac{\pi}{4}\!\cdot\!\tan x}{\tan\frac{\pi}{4} - \tan x} \;=\;\frac{1 + \tan x}{1 - \tan x}\)
\(\displaystyle \text{then: }\;\frac{\tan(2x)}{\cot\left(\frac{\pi}{4}-x\right)} \;\;=\;\;\frac{\frac{2\tan x}{1-\tan^2\!x}}{\frac{1+\tan x}{1-\tan x}} \;\;=\;\;\frac{2\tan x}{(1-\tan x)(1+\tan x)}\cdot\frac{1-\tan x}{1+\tan x} \;\;=\;\;\frac{2\tan x}{(1+\tan x)^2}\)
\(\displaystyle \text{Therefore: }\;\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2} \;\;=\;\;\frac{2\cdot 1}{(1+1)^2} \;\;=\;\;\frac{2}{4}\;\;=\;\;\frac{1}{2}\)