1.

Evaluate limit x tends to pi/6 root 3 sinx - cosx/x - pi/6

Answer»

Given, \bold{ \lim_{x \to\ {\pi/6}} \frac{\sqrt{3}sinx-cosx}{x-\pi/6}}
First of all we have to CHECK form of the LIMIT ,
PUT x = π/6 ,
(√3 sinπ/6 - cosπ/6)(π/6 - π/6) = 0/0 is the form of limit
We know, asinA -bcosA = √(a² + b²)sin{A - tan⁻¹(b/a)}, use it here,
Then, √3sinx - cosx = √{√3² + 1²}sin{x - tan⁻¹(1/√3)}
= 2sin(x - π/6)
Now, limit converts in \bold{\lim_{x\to{\pi/6}}\frac{2sin(x-\pi/6)}{x-\pi/6}}

Use the standard form for SOLUTION of Limit ,
Lim_{f(x)→f(a) sin{f(x)-f(a)}/{f(x) - f(a)} = 1 , use this here ,
Then, \bold{\lim_{x\to{\pi/6}}\frac{2sin(x-\pi/6)}{x-\pi/6}}
\bold{\lim_{x\to{\pi/6}}\frac{2sin(x-\pi/6)}{x-\pi/6}} = 2 × 1 = 2


HENCE, answer is 2


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