[![enter image description here](https://i.stack.imgur.com/7KedG.png)](https://i.stack.imgur.com/7KedG.png)

**Note: required to use little o**

a. ($x^2-x^4/2 + o(x^4) -(1+x^2+x^4/2+o(x^4) +1)/ (1+x^4)^{1/7}-1 $

Simplifying: $-x^4+o(x^4)-o(x^4) / (1+x^{4/7}+o(x^4)-1$.
Taking limit we get:
$-7$.

Concerns/Confusions: Not sure what happens to little os when subtracted in numerator (know they should be different) (confused overall by this concept). Actually not sure how to get Taylor Series of $(1+x^4)^{1/7}$, had to Wolfram Alpha it. Could someone explain to me, how to derive it by hand. I know it has to do with the expansion of $(1+x)$, and we use $x^4$, but the $7^\text{th}$ root throws me off. I think calculating derivatives to find the coefficients would be too hard, but perhaps this may be the only way.

b.Apparently I need a change of variable to use Taylor's Theorem. So I set $y=1/x$.
I get: 

$\lim\_{y \to 0} 1/y +\sqrt{(1/y)^2+23/y}$
==========================================

=
$\sqrt{(1+23y)}+1$ /$(y)$. 

Expanding numerator:
$1+11.5y+o(y)+1$/ (y).

Taking limit as $x$ goes to $0$, we get $23/2$ or $11.5$.

The actual answer, however is $-23/2$, so I am confused. I think perhaps I should put a negative in front of the square root sign, to get the right answer? Not too sure, could really use help. And should I do the limit $\lim\_{y \to 0^-}$ instead of just $\lim\_{y \to 0}$. So approaching from left hand side?