find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit

My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\mathcal{O}(x^3)$$ 
Now plugging in $x^2$ for x we get $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)$$ 
and from the common Maclaurin polynomial we have that $e^t=1+t+\frac{t^2}{2}+ \mathcal{O}(t^3)$. Plugging in $\sqrt{x^2+1}$ for $t$ we get:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6) 
}$$
which in turn is: 
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6) 
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)$$ 
hence we have: 
$$\lim_{x\rightarrow 0} \frac{1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{}O(x^6)-a-bx^2}{x^4}$$
And i argue that $a=2, b=1/2$ and the $\lim=-1/8 $
However the book disagrees with me and argues that the they should be $a=e, b=e/2$ and limit $=0$ 
i mean i can see how theyd done it, by not expanding $e^t$ but by only expanding $\sqrt{x^2+1}$ however what i dont understand how come we didnt get the same values or at least the same value for the limit?

First rewrite the equation replacing $x$ by $X$: Thus we require $L=\lim_{X\rightarrow 0}\dfrac{e^{\sqrt{1+X^2}}-a-bX^2}{X^4}$
Now let $x:=\sqrt{1+X^2}$, then $X^2=x^2-1$ with $X^4=(x^2-1)^2$ and $x\rightarrow 1$ as $X\rightarrow 0$.
Now, we have $L=\lim_{x\rightarrow 1}\dfrac{e^x-a-b(x^2-1)}{(x^2-1)^2}=\dfrac{\lim_{x\rightarrow 1}e^x-a-b(x^2-1)}{\lim_{x\rightarrow 1}(x^2-1)^2}=\dfrac{e-a}{0}$
Thus $L$ is not defined (finite) unless $e-a=0$. Now if $a=e$ we have $L=\frac{0}{0}$, so we may use L'Hopital.
Consider then $L=\dfrac{\lim_{x\rightarrow 1}e^x-2bx}{\lim_{x\rightarrow 1}4(x^2-1)x}=\dfrac{e-2b}{0}$
which is not defined unless $e-2b=0$. 
Thus for a finite limit it is necessary that $b=\frac{e}{2}$ and $a=e$. 
Again use L'Hopital:
$L=\dfrac{\lim_{x\rightarrow 1}e^x-e}{\lim_{x\rightarrow 1}12x^2-4}=\dfrac{e-e}{8}=0$ as required.
