What is the limit of the following problem? 
I need to find the limit of the following function :
$$\lim_{x\rightarrow 0} \  \frac{a^{\ x} \ b^{ \ x} - b^{ \ x} - a^{\ x} +1}{x^{ \ 2}}$$

I know that it is in $\frac{0}{0}$ form so I have applied L'Hospital rule as seen below:
$$\lim_{x\rightarrow 0}\frac{(ab)^{x} \ log(b) + (ab)^{x} \ log(a) - b^x \ log(b) - a^x \ log(a)}{2x} $$
Applying the L'Hospital rule again I ended up with:
$$\lim_{x\rightarrow 0}\frac{(ab)^{x}\log(ab) \left [ log(a) + log(b) \right ]\ - \ b^{x} \ (log \ b)^{2} \ - a^{x}\ (log \ a)^{2}}{2} \  $$
$$ =\frac{(log \ a)^{2} + (log \ b)^{2} + 2log(a)log(b) \ - (log \ a)^{2} - (log \ b)^{2}}{2}\ $$
$$=log(a)log(b)$$
Is the answer and method correct? Also if correct, is there a shorter way of solving it?

The method is correct, but you can observe that
$$
a^xb^x-a^x-b^x+1=(a^x-1)(b^x-1)
$$
and so you just need to compute
$$
\lim_{x\to0}\frac{a^x-1}{x}
$$
which is the derivative at zero of $f(x)=a^x$, which is well known to be $\log a$. Thus
$$
\lim_{x\to0}\frac{a^x-1}{x}\frac{b^x-1}{x}=(\log a)(\log b)
$$
