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?
 A: Hint. When you write
$$
e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6) 
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)
$$ it is wrong, since it is not right that, as $x \to 0$,
$$
\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)^m=\mathcal{O}(x^6),\qquad m=2,3,\cdots.
$$
Please look at
$$
\lim_{x \to 0}\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)^m.
$$
You can't neglect the previous quantities.
Do you see the issue?
A: 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 $\lim = 0$.
A: Simplified computation:
Let $\sqrt{1+x^2}=t+1$, so that
$$\lim_{x\to0}\frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}=\lim_{t\to0}\frac{e\cdot e^t-a-bt(2+t)}{t^2(2+t)^2}\\
=\lim_{t\to0}\frac{e+et+\dfrac{et^2}2+\dfrac{et^3}{3!}+\cdots-a-2bt-bt^2}{4t^2}.$$
Then it is obvious that we need 
$$a=e,b=\frac e2$$ and the third coefficient is
$$\frac e2-\frac e2=0.$$
A: 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 $0$.
A: 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.
