Trigonometric Limit Problem

  • Thread starter terryds
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Homework Statement


[/B]
##\lim x\rightarrow \frac{\pi }{4} (\frac{1-\tan x}{\sin x - \cos x})##

The Attempt at a Solution


[/B]
By assuming y = x-π/4 , the limit become :

##
\lim y\rightarrow 0 (\frac{1- \tan (y+\frac{\pi}{4})}{\sin (y+\frac{\pi}{4}) - \cos (y+\frac{\pi}{4})})
= \lim y\rightarrow 0 (\frac{1- (y + (\frac{\pi}{4})) (\frac{\tan (y+\frac{\pi}{4})}{y+(\frac{\pi}{4})})}{(y + (\frac{\pi}{4})) \frac{\sin (y+\frac{\pi}{4})}{(y + (\frac{\pi}{4}))} - \cos (y+\frac{\pi}{4})})
= (\frac{1-\frac{\pi}{4}(1)}{\frac{\pi}{4}(1)-\frac{\sqrt{2}}{2}})##

But, using the identity tan x = sin x/ cos x and by graph, I get the answer is ##-\sqrt{2}##

So, please tell me the wrong that I did..
Why can't we just use the theorem lim x-> 0 tan x/x = 1 and lim x->0 sin x/x =1 ??
I don't understand the steps to solve a trigonometric limit, because using different methods, the answer can be different
 

Answers and Replies

  • #2
Orodruin
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Why can't we just use the theorem lim x-> 0 tan x/x = 1 and lim x->0 sin x/x =1 ??
Because the argument of the tangent is not what is approaching zero. You need the Taylor expansion of the expressions around y=0, not x = 0.
 
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  • #3
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I don't understand the steps to solve a trigonometric limit, because using different methods, the answer can be different
No. Using a different method should not result in a different limit value. If you use two different methods to evaluate a limit, and get two different answers, then you have made a mistake in one (at least) of those methods.
 
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