Question

Lim x tends to pi/6 2sinx-1 /pi-6x

Answer 1

SOLUTION

TO EVALUATE

$\displaystyle \sf{\lim_{x \to \frac{\pi}{6} } \: \frac{2 \sin x - 1}{\pi - 6x} }$

EVALUATION

SOLVE USING L'HOSPITAL RULE

$\displaystyle \sf{\lim_{x \to \frac{\pi}{6} } \: \frac{2 \sin x - 1}{\pi - 6x} } \: \: \: \: \: \: \: \: \: \: \: \bigg( \: \frac{0}{0} \: \: form \: \bigg)$

$\displaystyle \sf{ = \lim_{x \to \frac{\pi}{6} } \: \frac{2 \cos x }{- 6} }$

$\displaystyle \sf{ =\: \frac{2 \cos \frac{\pi}{6} }{- 6} }$

$\displaystyle \sf{ =\: \frac{2 \times \frac{ \sqrt{3} }{2} }{- 6} }$

$\displaystyle \sf{ =\: - \frac{ \sqrt{3} }{6} }$

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Answer 2

Given :   Lim x→π/6      (2sinx - 1) /(π - 6x)

To Find :  Solve

Solution:

Lim x→π/6      (2sinx - 1) /(π - 6x)

Substituting x = π/6

2sinx - 1 = 2sin(π/6) - 1   = 2(1/2) - 1 = 0

π - 6x = π - 6(π/6)  = 0

0/0 form

Hence apply L' Hosp rule

Take derivative of numerator and denominator separately

Lim x→π/6      (2cosx) /(-6)

=> Lim x→π/6      cosx  /(-3)

Substituting x = π/6

=    cosπ/6  /(-3)

= (√3 / 2) / (-3)

=  -√3 / 6

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