Question

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

Answer 1

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

Answer 2

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