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?

Consider $f(x)=e^{1+x}$. By your reasoning, since $1+x=1+x+O(x^2)$, you get that:
$$f(x)=1+(1+x)+O(x^2)$$
But that isn't correct. If it was correct, then $f(0)=2$, but we know $f(0)=e$. 
Now, if you did the full substitution you'd get:
$$f(x)=1+(1+x)+\frac{(1+x)^2}{2!}+\frac{(1+x)^3}{3!}+\cdots$$
The problem is that the terms $\frac{(1+x)^k}{k!}$ keep adding to the constant and linear terms.
What you really should get is $$e^{1+x+O(x^2)}=e^{1}\left(1+\left(x+O(x^2)\right)+\frac{\left(x+O(x^2)\right)^2}{2!}+\cdots\right)$$
Now, $\left(x+O(x^2)\right)^2=O(x^2)$. So you get:
$$e^{1+x}=e+ex+O(x^2)$$
In your case, you get:
$$\begin{align}e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6) 
}&=e\left(1+\left(\frac{x^2}{2}-\frac{x^4}{8}\right)+\frac{\left(\frac{x^2}{2}-\frac{x^4}{8}\right)^2}{2!}+O(x^6)\right)\\
&=e\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^4}{8}+O(x^6)\right)\\
&=e+\frac{e}{2}x^2+O(x^6)
\end{align}$$
And you get $a=e,b=\frac{e}{2},$ and $\lim = 0$.
