How is this equal to one?

Question

√(x+Sx)-√x

lim_Sx→0 sin ( 2 ) = 1 (How is this equal to one?)

√(x+Sx)-√x

2

AND

√(x+Sx)+√x

lim_Sx→0 sin ( 2 ) = -1 sin√x

√(x+Sx)+√x 2√x

2

Answer 1

We know that sinx = x - x³/6 + x⁵ /120 +.....

So that from given question let t = { √(x+Sx) - √x} / 2

Then we can write question as lim_sx-0 sint / t.

Expand the sin then lim_t-0( t - t³/6 + t⁵/120+...) / t

= lim_t-0 (1 - t²/6 + t⁴/120+...)

Put t = 0 then = 1- 0²_/6 + 0⁴/120+..

= 1. It's proved

Answer 2

https://drive.google.com/file/d/1KI2i6ZptncZuCVAmyWenM0S4wxr2yGxV/view?usp=sharing