Explain the expression that sqrt(1+x) is "close" to 1+(1/2)*x for "small" x.
I tried to solved with epsilon-delta:
Let e>0, so there is delta(e)>0 so that
|x|<delta ==> |sqrt(1+x)-{1+(1/2)*x}|<e
Can anybody help me please...?
Looking to all three factors of the Taylor serie
$\displaystyle f^{(n)}(x=0)$ will become positive and negative alternatly, but the absolute value will become lower for higher n.
$\displaystyle x^n$ will become lower for higher n (for small x)
$\displaystyle 1/n!$ will become lower for higher n.
As you can see, all the three terms becomes lower for higher n. This means that the lower n terms will become dominant. The smaller x is, the more dominant the first terms will be.