Question

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. lim (tan(4x))^x x→0+

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:

\(1\)