How do you find the linearization of $sqrt(7+2x)$ at the point a=0?

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Answer 1 · 2017-11-25T16:09:41

The answer is $=sqrt7(1+1/7x-1/98x^2+1/686x^3+o(x^3))$

We need

$(1+x)^n=1+n/(1!)x+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+o(x^3)$

Therefore,

$sqrt(7+2x)=sqrt7(1+2/7x)^(1/2)$

$=sqrt7(1+1/2*2/7x+((1/2)(-1/2))/2*(2/7)^2x^2+((1/2(-1/2)(-3/2)))/(6)*(2/7x)^3+o(x^3))$

$=sqrt7(1+1/7x-1/98x^2+1/686x^3+o(x^3))$

How do you find the linearization of sqrt(7+2x)sqrt(7+2x) at the point a=0?