A Horrible looking limit I have the following limit question:
$$\lim_{x \rightarrow 1 }\frac {({\rm log} (1+x)-{\rm log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\rm sin}\space \pi x}$$
This has the form $\frac {0}{0}$ and hence I can apply L'Hospital's Rule. But, this is too large and that way it can go terribly wrong. Is there any  clever way to do this? 
I found it in a MCQ test and it has the following four alternatives: 
A. $\frac{9}{\pi}{\rm log}\space \frac{4}{e}$
B. 1
C. $\frac{3}{\pi}{\rm log} \space \frac{2}{e}$
D. $\frac{1}{\pi}$.

Hint Apply L'H to both
$$\lim_{x \rightarrow 1 }\frac {{\rm log} (1+x)-{\rm log}\space 2}{{\rm sin}\space \pi x}  $$
and
$$\lim_{x \rightarrow 1 }\frac {4^{x-1}-1}{(7+x)^{1/3}-(1+3x)^{1/2}} $$
