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

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.\n\nlim x→∞ x^(ln(6))/(1 + ln(x))

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.\n\nlim x→∞ x^(ln(6))/(1 + ln(x))

Answer

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)