question 4. 1 point\nwhat is lim┬(x→1)⁡h(x)?\nh(x)=(x² - x)/ln x\nprovide your answer below:

Answer

Explanation:

Step1: Check form of limit

When (x\rightarrow1), (\lim_{x\rightarrow1}\frac{x^{2}-x}{\ln x}) is in the (\frac{0}{0}) form since (1^{2}-1 = 0) and (\ln(1)=0).

Step2: Apply L - H rule

L'Hopital's rule states that if (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}) is in the (\frac{0}{0}) or (\frac{\infty}{\infty}) form, then (\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}). Differentiate (f(x)=x^{2}-x) and (g(x)=\ln x). (f'(x)=2x - 1) and (g'(x)=\frac{1}{x}).

Step3: Calculate new limit

(\lim_{x\rightarrow1}\frac{2x - 1}{\frac{1}{x}}=\lim_{x\rightarrow1}(2x - 1)x).

Step4: Substitute (x = 1)

((2\times1-1)\times1=1).