Answer:
1/3
Step-by-step explanation:
The ratio is undefined at x=0, so we presume that's where we're interested in the limit. Both numerator and denominator are zero at x=0, so L'Hôpital's rule applies. According to that rule, we replace numerator and denominator with their respective derivatives.
[tex]\displaystyle\lim\limits_{x\to 0}\dfrac{\tan{(4x)}}{4\tan{(3x)}}=\lim\limits_{x\to 0}\dfrac{\tan'{(4x)}}{4\tan'{(3x)}}=\lim\limits_{x\to 0}\dfrac{4\sec{(4x)^2}}{12\sec{(3x)^2}}=\dfrac{4}{12}\\\\\boxed{\lim\limits_{x\to 0}\dfrac{\tan{(4x)}}{4\tan{(3x)}}=\dfrac{1}{3}}[/tex]
