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Sagot :
Certainly! Let's evaluate the limit:
[tex]\[ \lim_{x \to \frac{\pi}{6}} \left( \frac{2 \tan x - 2}{2 \cos x - \sqrt{3}} \right). \][/tex]
### Step-by-Step Solution:
1. Substitute [tex]\(x\)[/tex] with [tex]\(\frac{\pi}{6}\)[/tex] in the expression:
For [tex]\(\tan\left(\frac{\pi}{6}\right)\)[/tex] and [tex]\(\cos\left(\frac{\pi}{6}\right)\)[/tex], we use known trigonometric values.
- [tex]\(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)[/tex]
- [tex]\(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)[/tex]
2. Evaluate the expression:
Substitute these values into the given expression:
[tex]\[ \frac{2 \tan \left( \frac{\pi}{6} \right) - 2}{2 \cos \left( \frac{\pi}{6} \right) - \sqrt{3}} = \frac{2 \cdot \frac{1}{\sqrt{3}} - 2}{2 \cdot \frac{\sqrt{3}}{2} - \sqrt{3}} \][/tex]
3. Simplify the numerator and the denominator:
[tex]\[ \text{Numerator: } 2 \cdot \frac{1}{\sqrt{3}} - 2 = \frac{2}{\sqrt{3}} - 2 \][/tex]
Bring the terms to a common denominator:
[tex]\[ \frac{2}{\sqrt{3}} - 2 = \frac{2 - 2\sqrt{3}}{\sqrt{3}} \][/tex]
[tex]\[ = \frac{2 - 2\sqrt{3}}{\sqrt{3}} \][/tex]
[tex]\[ = 2 \left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right) \][/tex]
[tex]\[ \text{Denominator: } 2 \cdot \frac{\sqrt{3}}{2} - \sqrt{3} = \sqrt{3} - \sqrt{3} = 0 \][/tex]
4. Form of the limit:
The expression now looks like:
[tex]\[ \frac{2 \left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right)}{0} \][/tex]
The presence of [tex]\(0\)[/tex] in the denominator suggests that the limit approaches infinity or negative infinity, depending on the sign of the numerator.
- Since [tex]\(2 \left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right)\)[/tex] is a non-zero finite value and [tex]\((1 - \sqrt{3}) < 0\)[/tex], meaning this fraction's value is negative.
- However, when dealing with limits and approaching the value of [tex]\(0\)[/tex] from a point where numerator is non-zero, the general behavior of such functions generally implies that the limit is [tex]\( \infty \)[/tex] or [tex]\(-\infty\)[/tex].
### Conclusion:
Thus, the limit as [tex]\( x \to \frac{\pi}{6} \)[/tex] of the given function is undefined due to the singularities that make the function approach infinity.
[tex]\(\lim_{x \to \frac{\pi}{6}}\left(\frac{2 \tan x-2}{2 \cos x-\sqrt{3}}\right) = \boxed{\infty}\)[/tex]
Therefore the limit [tex]\( \lim_{x \to \frac{\pi}{6}} \left( \frac{2 \tan x - 2}{2 \cos x - \sqrt{3}} \right) \)[/tex] is indeed [tex]\( \infty \)[/tex].
[tex]\[ \lim_{x \to \frac{\pi}{6}} \left( \frac{2 \tan x - 2}{2 \cos x - \sqrt{3}} \right). \][/tex]
### Step-by-Step Solution:
1. Substitute [tex]\(x\)[/tex] with [tex]\(\frac{\pi}{6}\)[/tex] in the expression:
For [tex]\(\tan\left(\frac{\pi}{6}\right)\)[/tex] and [tex]\(\cos\left(\frac{\pi}{6}\right)\)[/tex], we use known trigonometric values.
- [tex]\(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)[/tex]
- [tex]\(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)[/tex]
2. Evaluate the expression:
Substitute these values into the given expression:
[tex]\[ \frac{2 \tan \left( \frac{\pi}{6} \right) - 2}{2 \cos \left( \frac{\pi}{6} \right) - \sqrt{3}} = \frac{2 \cdot \frac{1}{\sqrt{3}} - 2}{2 \cdot \frac{\sqrt{3}}{2} - \sqrt{3}} \][/tex]
3. Simplify the numerator and the denominator:
[tex]\[ \text{Numerator: } 2 \cdot \frac{1}{\sqrt{3}} - 2 = \frac{2}{\sqrt{3}} - 2 \][/tex]
Bring the terms to a common denominator:
[tex]\[ \frac{2}{\sqrt{3}} - 2 = \frac{2 - 2\sqrt{3}}{\sqrt{3}} \][/tex]
[tex]\[ = \frac{2 - 2\sqrt{3}}{\sqrt{3}} \][/tex]
[tex]\[ = 2 \left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right) \][/tex]
[tex]\[ \text{Denominator: } 2 \cdot \frac{\sqrt{3}}{2} - \sqrt{3} = \sqrt{3} - \sqrt{3} = 0 \][/tex]
4. Form of the limit:
The expression now looks like:
[tex]\[ \frac{2 \left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right)}{0} \][/tex]
The presence of [tex]\(0\)[/tex] in the denominator suggests that the limit approaches infinity or negative infinity, depending on the sign of the numerator.
- Since [tex]\(2 \left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right)\)[/tex] is a non-zero finite value and [tex]\((1 - \sqrt{3}) < 0\)[/tex], meaning this fraction's value is negative.
- However, when dealing with limits and approaching the value of [tex]\(0\)[/tex] from a point where numerator is non-zero, the general behavior of such functions generally implies that the limit is [tex]\( \infty \)[/tex] or [tex]\(-\infty\)[/tex].
### Conclusion:
Thus, the limit as [tex]\( x \to \frac{\pi}{6} \)[/tex] of the given function is undefined due to the singularities that make the function approach infinity.
[tex]\(\lim_{x \to \frac{\pi}{6}}\left(\frac{2 \tan x-2}{2 \cos x-\sqrt{3}}\right) = \boxed{\infty}\)[/tex]
Therefore the limit [tex]\( \lim_{x \to \frac{\pi}{6}} \left( \frac{2 \tan x - 2}{2 \cos x - \sqrt{3}} \right) \)[/tex] is indeed [tex]\( \infty \)[/tex].
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