find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…

Explanation:

Step1: Take the natural logarithm

Let ( y = (\tan(4x))^{x}). Then (\ln y=x\ln(\tan(4x))=\frac{\ln(\tan(4x))}{\frac{1}{x}}). As (x\to0^{+}), (\tan(4x)\to0), so (\ln(\tan(4x))\to-\infty) and (\frac{1}{x}\to+\infty). This is (\frac{-\infty}{\infty}) form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and denominator. The derivative of (\ln(\tan(4x))) is (\frac{4\sec^{2}(4x)}{\tan(4x)}=\frac{4}{\sin(4x)\cos(4x)}), and the derivative of (\frac{1}{x}) is (-\frac{1}{x^{2}}). So (\lim_{x\to0^{+}}\frac{\ln(\tan(4x))}{\frac{1}{x}}=\lim_{x\to0^{+}}\frac{\frac{4}{\sin(4x)\cos(4x)}}{-\frac{1}{x^{2}}}=\lim_{x\to0^{+}}\frac{-4x^{2}}{\sin(4x)\cos(4x)}). Since (\sin(4x)\cos(4x)=\frac{1}{2}\sin(8x)), then (\lim_{x\to0^{+}}\frac{-4x^{2}}{\frac{1}{2}\sin(8x)}=\lim_{x\to0^{+}}\frac{-8x^{2}}{\sin(8x)}). This is (\frac{0}{0}) form. Apply L'Hospital's Rule again. The derivative of (- 8x^{2}) is (-16x), and the derivative of (\sin(8x)) is (8\cos(8x)). So (\lim_{x\to0^{+}}\frac{-16x}{8\cos(8x)} = 0).

Step3: Find the original limit

Since (\lim_{x\to0^{+}}\ln y = 0), then (\lim_{x\to0^{+}}y=e^{0})

Answer:

(1)