Question

plz..solve it guys..​

Answer 1

Answer:

Given, \bold{ \lim_{x \to\ {\pi/6}} \frac{\sqrt{3}sinx-cosx}{x-\pi/6}}lim

x→ π/6

x−π/6

3

sinx−cosx

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}}lim

x→π/6

x−π/6

2sin(x−π/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}}lim

x→π/6

x−π/6

2sin(x−π/6)

\bold{\lim_{x\to{\pi/6}}\frac{2sin(x-\pi/6)}{x-\pi/6}}lim

x→π/6

x−π/6

2sin(x−π/6)

= 2 × 1 =2

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Answer 2

$\huge\red{Hello\:Buddy}$

hope it helps