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## Homework Statement

Use Taylor's Theorem to show that

√(x+1)=1+(1/2)x+O(x

^{2})

for x sufficiently small.

Here's what I did:

f(x)= √x+1

f'(x)= (1/2)(x+1)

^{(-1/2)}

Then using x

_{0}=0,

f(0)= 1, f'(0)=1/2.

√x+1=1+(1/2)x-(1/8)x

^{2}(cx+1)

^{(-3/2)}

So, then using h as a parameter:

l√(h+1) -1-(1/2)h l ≤ 1/8(h

^{2})

Finally,

√(h+1) = 1+(1/2)h+O(h

^{2})

Is this correct?

I' m having difficulty understanding the meaning of O, can someone please explain in simple terms?

Thank you.

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