Question

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

lim x→∞ x^(ln(6))/(1 + ln(x))

Explanation:

Step1: Let \( y = x^{\frac{\ln(6)}{1+\ln(x)}}\)

Take the natural logarithm of both sides: \(\ln y=\frac{\ln(6)\ln x}{1 + \ln x}\)

Step2: Find the limit of \(\ln y\) as \(x\to\infty\)

\(\lim_{x\to\infty}\ln y=\lim_{x\to\infty}\frac{\ln(6)\ln x}{1+\ln x}\)
Let \(t = \ln x\), as \(x\to\infty\), \(t\to\infty\). Then \(\lim_{t\to\infty}\frac{\ln(6)t}{1 + t}\)
Divide numerator and denominator by \(t\): \(\lim_{t\to\infty}\frac{\ln(6)}{\frac{1}{t}+ 1}\)
Since \(\lim_{t\to\infty}\frac{1}{t}=0\), we have \(\lim_{t\to\infty}\frac{\ln(6)}{\frac{1}{t}+ 1}=\ln(6)\)

Step3: Find the limit of \(y\)

Since \(\lim_{x\to\infty}\ln y=\ln(6)\), and \(y = e^{\ln y}\), then \(\lim_{x\to\infty}y = e^{\ln(6)}\)

Answer:

\(6\)