- #1
necessary
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:!)
lim(1-x).tan Pi X/2=?
x->1
please help meee :!)
lim(1-x).tan Pi X/2=?
x->1
please help meee :!)
Find the Limit of (1-x)tan(πx/2) as x Approaches 1 - Get Help Here!
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SUMMARY
The limit of (1-x)tan(πx/2) as x approaches 1 can be evaluated by substituting y = 1 - x and using the identity tan(θ) = sin(θ)/cos(θ). This approach simplifies the expression, leading to a clear resolution. Additionally, the limit of (1 - sin(x))/cos(x) as x approaches π/2 can be tackled by multiplying both the numerator and denominator by (1 + sin(x)), which further simplifies the limit calculation. These methods provide effective strategies for solving trigonometric limits.
PREREQUISITES- Understanding of limits in calculus
- Familiarity with trigonometric identities
- Knowledge of L'Hôpital's Rule for indeterminate forms
- Basic algebraic manipulation skills
- Study L'Hôpital's Rule for evaluating indeterminate limits
- Learn about trigonometric limit properties and techniques
- Explore the use of Taylor series for approximating trigonometric functions
- Practice solving limits involving sin and cos functions in calculus
Students studying calculus, mathematics educators, and anyone seeking to improve their skills in evaluating limits involving trigonometric functions.
- #2
mathman
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Let y=1-x and use tan=sin/cos and also sinu=cos(pi/2 -u), you will get your answer easily.
- #3
necessary
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thanks..
one more guestion that I can't solve is...lim 1-sinX/cosx =?
X->pi/2
(I don't like trigonometric questions in limit)
thanks:)
one more guestion that I can't solve is...lim 1-sinX/cosx =?
X->pi/2
(I don't like trigonometric questions in limit)
thanks:)
- #4
VietDao29
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\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}
You can try multiplying both numerator and denominator by (1 + sin x), something like:
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{(1 - \sin x)(1 + \sin x)}{\cos x(1 + \sin x)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin ^ 2 x}{\cos x(1 + \sin x)}
1 - sin2x = ...
Can you go from here?
----------------------
Or you can try a different way:
1 - \sin x = \sin \left( \frac{\pi}{2} \right) - \sin x = 2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)
\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\sin\left( \frac{\pi}{2} - x \right)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2\sin \left( \frac{\pi}{4} - \frac{x}{2} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{ \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}
Viet Dao,
You can try multiplying both numerator and denominator by (1 + sin x), something like:
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{(1 - \sin x)(1 + \sin x)}{\cos x(1 + \sin x)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin ^ 2 x}{\cos x(1 + \sin x)}
1 - sin2x = ...
Can you go from here?
----------------------
Or you can try a different way:
1 - \sin x = \sin \left( \frac{\pi}{2} \right) - \sin x = 2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)
\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\sin\left( \frac{\pi}{2} - x \right)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2\sin \left( \frac{\pi}{4} - \frac{x}{2} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{ \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}
Viet Dao,
- #5
necessary
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thanks,yu are a good man 
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