Determine the values of $a$ and $b$ if
$$ \sqrt{x^2 +1}=ax + b +o(x) $$
when $x\to 0$ and prove the result, where $o(x)$ corresponds to the notation little $o$.
My first attempt was to rewrite the equation as
$$ \sqrt{x^2 +1} -ax - b = o(x) $$
and use the definition of little $o$, then I would need to show that the next lim is equal to zero
$$ \lim_{x \to 0} \frac{\sqrt{x^2 +1} -ax - b }{x} = 0$$
in order to use L'Hopital I need $0/0$ in the direct evaluation. So if $f(x) = \sqrt{x^2 + 1} -ax -b$ then I need $0= f(0) = 1-b $ or equivalent $b = 1$.
Using L'Hopital we need
$$ \frac{ \lim_{x \to 0} \frac{x-a}{\sqrt{x^2+1}} }{\lim_{x \to 0} 1} = 0$$
Substituting x,
$$ \lim_{x \to 0} \frac{ -a }{1} = 0$$
So $a=0$. But I'm not sure I'm susing correctly the implications in L'Hopital Theorem cause I known the limit exists and I would like L'Hopital to be true but I feel I'm forcing the hyptesis. Is there anoher way t find $a$ and $b$?
There is simpler, set $f(x)=\sqrt{x^2+1}$ which is derivable in $0$.
Notice $f(0)=1$ therefore $\dfrac{f(x)-f(0)}{x}\to f'(0)=0$
Which is nothing more than $\sqrt{1+x^2}=1+o(x)$