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

Explanation:

Step1: Let ( y=(3x + 1)^{\cot(x)} )

Take the natural logarithm of both sides: ( \ln y=\cot(x)\ln(3x + 1)=\frac{\ln(3x + 1)}{\tan(x)} )

Step2: Find the limit of ( \ln y ) as ( x\to0^{+} )

As ( x\to0^{+} ), we have the indeterminate form ( \frac{0}{0} ). Apply L'Hospital's Rule. Differentiate the numerator and denominator: The derivative of ( \ln(3x + 1) ) is ( \frac{3}{3x+1} ), and the derivative of ( \tan(x) ) is ( \sec^{2}(x) ) So ( \lim_{x\to0^{+}}\ln y=\lim_{x\to0^{+}}\frac{\frac{3}{3x + 1}}{\sec^{2}(x)} )

Step3: Evaluate the limit of ( \ln y )

Substitute ( x = 0 ) into ( \frac{\frac{3}{3x + 1}}{\sec^{2}(x)} ) ( \frac{\frac{3}{3\times0+1}}{\sec^{2}(0)}=\frac{3}{1}=3 )

Step4: Find the limit of ( y )

Since ( \lim_{x\to0^{+}}\ln y = 3 ), and ( y = e^{\ln y} ) By the continuity of the exponential - function ( \lim_{x\to0^{+}}y=\lim_{x\to0^{+}}e^{\ln y}=e^{\lim_{x\to0^{+}}\ln y} )

Answer:

( e^{3} )