How Do You Solve This Trigonometric Limit Problem?

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SUMMARY

The discussion focuses on solving the trigonometric limit problem: lim x→π/4 (1−tan x)/(sin x−cos x). The correct approach involves substituting y = x−π/4 and applying Taylor expansions around y=0. The final limit evaluates to −√2, contrasting with incorrect assumptions using standard limits like lim x→0 tan x/x = 1 and lim x→0 sin x/x = 1. The key takeaway is that using Taylor series is essential for accurate limit evaluation in this context.

PREREQUISITES
  • Understanding of trigonometric functions and their limits
  • Familiarity with Taylor series expansions
  • Knowledge of limit theorems in calculus
  • Ability to manipulate algebraic expressions involving trigonometric identities
NEXT STEPS
  • Study Taylor series expansions for trigonometric functions
  • Practice solving limits involving trigonometric identities
  • Explore advanced limit theorems and their applications
  • Review common mistakes in limit evaluation and how to avoid them
USEFUL FOR

Students studying calculus, particularly those focusing on limits and trigonometric functions, as well as educators looking for examples of limit evaluation techniques.

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
 
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terryds said:
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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terryds said:
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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