Compute $\lim_{x \to 0} \frac{\sin(x)+\cos(x)-e^x}{\log(1+x^2)}$. Question:

Compute $\lim_{x \to 0} \frac{\sin(x)+\cos(x)-e^x}{\log(1+x^2)}$.

Attempt:
Using L'Hopital's Rule, I have come to
$$ \lim_{x \to 0} \frac{\cos(x)}{2x} - \lim_{x \to 0} \frac{\sin(x)}{2x} - \lim_{x \to 0} \frac{e^x}{2x}.$$
My thought was to use the power series representations of these functions. However, that' doesn't seem to get me anywhere. Am I on the wrong track using L'Hopital's?
 A: Tricky computation:
\begin{eqnarray*}
\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})} &=&\frac{(\sin (x)-x)+(\cos
(x)-1)-(e^{x}-1-x)}{\log (1+x^{2})} \\
&=&\frac{\frac{(\sin (x)-x)}{x^{2}}+\frac{(\cos (x)-1)}{x^{2}}-\frac{%
(e^{x}-1-x)}{x^{2}}}{\frac{\log (1+x^{2})}{x^{2}}}.
\end{eqnarray*}
Now we use l'Hospital's rule to compute each of the four limits separatly
$\lim_{x\rightarrow 0}\frac{(\sin (x)-x)}{x^{2}}\overset{L^{\prime }HR}{=}%
\lim_{x\rightarrow 0}\frac{(\cos (x)-1)}{2x}\overset{L^{\prime }HR}{=}%
\lim_{x\rightarrow 0}\frac{-\sin (x)}{2}=\frac{-\sin (0)}{2}=0.$
$\lim_{x\rightarrow 0}\frac{(\cos (x)-1)}{x^{2}}\overset{L^{\prime }HR}{=}%
\lim_{x\rightarrow 0}\frac{-\sin (x)}{2x}\overset{L^{\prime }HR}{=}%
\lim_{x\rightarrow 0}\frac{-\cos (x)}{2}=\frac{-\cos (0)}{2}=-\frac{1}{2}$
$\lim_{x\rightarrow 0}\frac{(e^{x}-1-x)}{x^{2}}\overset{L^{\prime }HR}{=}%
\lim_{x\rightarrow 0}\frac{e^{x}-1}{2x}\overset{L^{\prime }HR}{=}%
\lim_{x\rightarrow 0}\frac{e^{x}}{2}=\frac{e^{0}}{2}=\frac{1}{2}$
$\lim_{x\rightarrow 0}\frac{\log (1+x^{2})}{x^{2}}\underset{x^{2}=y}{=}%
\lim_{y\rightarrow 0^{+}}\frac{\log (1+y)}{y}\overset{L^{\prime }HR}{=}%
\lim_{y\rightarrow 0^{+}}\frac{1/(1+y)}{1}=1.$ 
Therefore
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})} &=&\frac{%
\lim_{x\rightarrow 0}\frac{(\sin (x)-x)}{x^{2}}+\lim_{x\rightarrow 0}\frac{%
(\cos (x)-1)}{x^{2}}-\lim_{x\rightarrow 0}\frac{(e^{x}-1-x)}{x^{2}}}{%
\lim_{x\rightarrow 0}\frac{\log (1+x^{2})}{x^{2}}} \\
&=&\frac{0+(-\frac{1}{2})-(\frac{1}{2})}{1}=-1.
\end{eqnarray*}
A: Recall that, for $x$ near $0$, you have
$$\begin{align}
&e^x =1+x+\dfrac{x^2}{2}+\mathcal{O}(x^3)\\
&\cos x = 1-\dfrac{x^2}{2}+\mathcal{O}(x^3)\\
& \sin x = x+\mathcal{O}(x^3)\\
\end{align}
$$
 giving $$\cos x+\sin x-e^x =- x^2+\mathcal{O}(x^3)$$ 
Since
$$\begin{align}
&\ln (1+ x^2) = x^2+\mathcal{o}(x^3)
\end{align}
$$
then
$$\lim_{x \to 0} \frac{\sin(x)+\cos(x)-e^x}{\log(1+x^2)}=\lim_{x \to 0} \frac{-x^2}{x^2}=-1.$$
A: we get by the rule of L'Hospital
$$\lim_{x \to 0}\frac{\cos(x)-\sin(x)-e^x}{\frac{2x}{1+x^2}}$$
$$\lim_{ x\to 0}\frac{(1+x^2)(\cos(x)-\sin(x)-e^x}{2x}$$
$$\lim_{x \to 0}\frac{2(\cos(x)-\sin(x)-e^x)+2x(-\sin(x)-\cos(x)-e^x)}{2}=0$$
A: Here's how I would remake the question: $$lim \left( {cos x \over 2x}-{e^x \over2x} \right)- lim {sinx \over 2x}$$
Use L'hopital's rule for the first and you should get negative infinity. The value of the second is equal to 1/2, so the limit is negative infinity.
A: You made a mistake when you looked for derivative. When you apply L'Hopital's Rule you should get $$\lim_{x \to 0} \frac{\cos(x)-\sin(x)-e^x}{\frac{2x}{1+x^2}}$$ witch becomes 
$$\lim_{x \to 0} \frac{(x^2+1)(\cos(x)-\sin(x)-e^x)}{2x}$$ and when you apply L'Hopital's Rule again you should get $$\lim_{x \to 0} \frac{2x(\cos(x)-\sin(x)-e^x)+(1+x^2)(-\sin(x)-\cos(x)-e^x)}{2}=-1$$ Sorry for that mistake.
A: Computation with l'Hospital's rule no trickly:
\begin{eqnarray*}
&&\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})}\left. 
\overset{L^{\prime }HR}{=}\right. \lim_{x\rightarrow 0}\frac{\cos (x)-\sin
(x)-e^{x}}{\frac{2x}{1+x^{2}}} \\
&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \left. =\right. \lim_{x\rightarrow 0}\left( 1+x^{2}\right)
\lim_{x\rightarrow 0}\left( \frac{\cos (x)-\sin (x)-e^{x}}{2x}\right) 
\end{eqnarray*}
The first limit equals $+1.$ To compute the second limit we continue using
l'Hospital's rule:
$$
\overset{L^{\prime }HR}{=}\lim_{x\rightarrow 0}\left( \frac{-\sin (x)-\cos
(x)-e^{x}}{2}\right) =\frac{-0-1-1}{2}=-1.
$$
Thereore
$$
\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})}=1\times
(-1)=-1.
$$
Remark: If we have keeped the factor $1+x^{2}$ with the second one, the
computation would be much more complicated; indeed, this factor has a
contribution in complicating the computation. So putting it a side, it do
note creat trouble! 
If we have to keep it look the computation looks like this
\begin{eqnarray*}
&&\lim_{x\rightarrow 0}\frac{\sin (x)+\cos (x)-e^{x}}{\log (1+x^{2})}\left. 
\overset{L^{\prime }HR}{=}\right. \lim_{x\rightarrow 0}\frac{\cos (x)-\sin
(x)-e^{x}}{\frac{2x}{1+x^{2}}} \\
&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \left. =\right. \lim_{x\rightarrow 0}\frac{\left( 1+x^{2}\right)
\left( \cos (x)-\sin (x)-e^{x}\right) }{2x} \\
&&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \lim_{x\rightarrow 0}\frac{\cos
x-\sin x-e^{x}+x^{2}\cos x-x^{2}\sin x-x^{2}e^{x}}{2x} \\
&&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \lim_{x\rightarrow 0}\frac{-\sin
x-\cos x-e^{x}-e^{x}x^{2}-x^{2}\sin x-x^{2}\cos x-2e^{x}x-2x\sin x+2x\cos x}{%
2} \\
&&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \frac{-0-1-1-0-0-0-0-0+0}{2} \\
&&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. \frac{-2}{2} \\
&&\ \ \ \ \ \ \ \ \ \ \ \left. =\right. -1.
\end{eqnarray*}
