Question

Evaluate limx->a (sinx-sina)/(√x-√a)

Answer 1

lim [(sin x - sin a) /(√x - √a)] = 
x→a 

the limit is of the form 0/0 

let's apply the sum-to-product formula sin α - sin β = 2sin[(α - β)/2] cos[(α + β)/2]: 

lim { {2sin[(x - a)/2] cos[(x + a)/2]} /(√x - √a)] = 
x→a 

let's multiply numerator and denominator by √x + √a. yielding a difference of squares in the denominator: 

lim { {2(√x + √a) sin[(x - a)/2] cos[(x + a)/2]} /[(√x + √a)(√x - √a)]} = 
x→a 

lim { {2(√x + √a) sin[(x - a)/2] cos[(x + a)/2]} /[(√x)² - (√a)²]} = 
x→a 

lim { {2(√x + √a) sin[(x - a)/2] cos[(x + a)/2]} /(x - a)} = 
x→a 

let's rearrange this as: 

lim {(√x + √a) sin[(x - a)/2] cos[(x + a)/2] [2 /(x - a)]} = 
x→a 

lim { {(√x + √a) sin[(x - a)/2] cos[(x + a)/2]} /[(x - a)/2]} = 
x→a 

lim {(√x + √a) cos[(x + a)/2] { {sin[(x - a)/2]} /[(x - a)/2]} } = 
x→a 

where {sin[(x - a)/2]} /[(x - a)/2] is a notable limit of the form lim [(sin t) /t] = 1, 
................ ..................... .................... .................. ...............t→0 
in that, if we let (x - a)/2 = t, as x approaches "a", t tends to zero; 
then, letting x approach "a", we obtain: 

lim {[√(→a) + √a] cos{[(→a) + a]/2} (→1)} = 
x→a 

(√a + √a) cos[(a + a)/2] (1) = 

2(√a) cos[(2a)/2] = 


2(√a) cos a 

Answer 2

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