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. $\frac{1}{\pi}$.

You can write $\frac {({\rm log} (1+x)-{\rm log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\rm sin}\space \pi x}=\frac{\log(1+x)-\log(2)}{x-1}\frac{3 4^{x-1}-3x}{x-1}\frac{x-1}{(7+x)^{1/3}-(1+3x)^{1/2}}\frac{x-1}{sin(\pi x)}=\frac{f(x)-f(1)}{x-1}\frac{g(x)-g(1)}{x-1}\frac{x-1}{h(x)-h(1)}\frac{x-1}{k(x)-k(1)}$
with $f(x)=\log(1+x)$, $g(x)=3 4^{x-1}-3x$, $h(x)=(7+x)^{1/3}-(1+3x)^{1/2}$ and $k(x)=sin(\pi x)$. 
So $\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f'(1)=\frac{1}{2}$, $\lim_{x\to1}\frac{g(x)-g(1)}{x-1}=g'(1)=3(\ln(4)-1)$, $\lim_{x\to1}\frac{h(x)-h(1)}{x-1}=h'(1)=\frac{1}{3}7^{-2/3}-\frac{3}{4}$ and $\lim_{x\to1}\frac{k(x)-k(1)}{x-1}=k'(1)=\pi\cos(\pi)=-\pi$.
Since $h'(1)\neq 0$ and $k'(1)\neq 0$, Limit Laws give you that the limit you're looking for is $\frac{f'(1)g'(1)}{h'(1)k'(1)}$.  
PS: I don't know if $\log$ stands for logarithm in base $10$ or $e$ so I left as it is.
