Question

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

lim (3x + 1)^cot(x)
x→0+

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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:

\( e^{3} \)