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

Explanation:

Step1: Combine the fractions

First, combine the two fractions: $$\lim_{x\rightarrow1}\left(\frac{9x}{x - 1}-\frac{9}{\ln(x)}\right)=\lim_{x\rightarrow1}\frac{9x\ln(x)-9(x - 1)}{(x - 1)\ln(x)}$$ When (x = 1), the numerator (9x\ln(x)-9(x - 1)=9\times1\times\ln(1)-9(1 - 1)=0) and the denominator ((x - 1)\ln(x)=(1 - 1)\ln(1)=0). So, we can apply L'Hospital's Rule.

Step2: Apply L'Hospital's Rule once

Differentiate the numerator and the denominator. The derivative of the numerator (u = 9x\ln(x)-9(x - 1)): Using the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u = 9x), (v=\ln(x)), (u^\prime=9), (v^\prime=\frac{1}{x}), and ((-9(x - 1))^\prime=-9) (u^\prime=9\ln(x)+9x\times\frac{1}{x}-9=9\ln(x)) The derivative of the denominator (v=(x - 1)\ln(x)): Using the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u=x - 1), (v = \ln(x)), (u^\prime=1), (v^\prime=\frac{1}{x}) (v^\prime=\ln(x)+\frac{x - 1}{x}) So, (\lim_{x\rightarrow1}\frac{9x\ln(x)-9(x - 1)}{(x - 1)\ln(x)}=\lim_{x\rightarrow1}\frac{9\ln(x)}{\ln(x)+\frac{x - 1}{x}}) When (x = 1), the new numerator (9\ln(1)=0) and the new denominator (\ln(1)+\frac{1 - 1}{1}=0). So, we apply L'Hospital's Rule again.

Step3: Apply L'Hospital's Rule second - time

Differentiate the new numerator and denominator. The derivative of the numerator (y_1 = 9\ln(x)), (y_1^\prime=\frac{9}{x}) The derivative of the denominator (y_2=\ln(x)+\frac{x - 1}{x}=\ln(x)+1-\frac{1}{x}) (y_2^\prime=\frac{1}{x}+\frac{1}{x^{2}}) So, (\lim_{x\rightarrow1}\frac{\frac{9}{x}}{\frac{1}{x}+\frac{1}{x^{2}}}=\lim_{x\rightarrow1}\frac{9x}{x + 1})

Step4: Evaluate the limit

Substitute (x = 1) into (\frac{9x}{x + 1}) (\frac{9\times1}{1+1}=\frac{9}{2})

Answer:

(\frac{9}{2})