Answer:
1/16
Step-by-step explanation:
Place x=0 and you can see [tex]\frac{0}{0}[/tex] indefinitness. So you can apply the l'hosptial rule. Its basic you should
[tex]\lim_{x \to \ 0} \frac{g(x)}{f(x)} = \\[/tex] [tex]lim_{x \to \ 0} \frac{g'(x)}{f'(x)}[/tex] so apply the derivative
[tex]\frac{\frac{1}{x-4} +\frac{1}{\sqrt[4]{(1-x})^3 } }{2x}[/tex] and replace the x=0 and you'll see same answer [tex]\frac{0}{0}[/tex] re-apply the l'hospital rule and answer
[tex]\frac{\frac{-1}{(x-4)^2} + \frac{3}{16\sqrt[4]{(1-x)^7} } }{2}[/tex] and replace the 0 you can see the answer [tex]\frac{1}{16}[/tex]
