Question 989973
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Do you know and did you learn that


{{{sqrt(1 + x)}}} = {{{1 + (1/2)x}}} + the terms of lower orders of magnitude at   x ---> 0 ?


In any case,  it is so. 


Therefore,  {{{sqrt(1 + x)-1}}} = {{{(1/2)x}}} + the terms of lower orders of magnitude at   x ---> 0,     and further


 {{{(sqrt(1 + x)-1)/x}}} = {{{(1/2)}}} + the terms of lower orders of magnitude at   x ---> 0.


It implies that lim {{{(sqrt(1 + x)-1)/x}}} = {{{1/2}}} at   x ---> 0.