<s>[INST] 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?
 [/INST] Ignoring the squares, we have
$$\lim_{x\to0}\frac{e^{\sqrt{1+x}}-a-bx}{x^2}.$$
By adjusting $a$ and $b$, we can certainly get rid of the linear terms and the limit will exist.
So let's compute the Taylor development of $f(x):=e^{\sqrt{1+x}}$.
$$f(0)=e=a,$$
$$f'(x)=\frac{e^{\sqrt{1+x}}}{2\sqrt{1+x}},f'(0)=\frac e2=b,$$
$$f''(x)=\frac{e^{\sqrt{1+x}}}{4(1+x)}-\frac{e^{\sqrt{1+x}}}{4(1+x)^{3/2}},f''(0)=0.$$
Hence the limit is $0$.
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