Question

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

Explanation:

Step1: Check form of limit

When \(x
ightarrow1\), \(\lim_{x
ightarrow1}\frac{x^{2}-x}{\ln x}\) 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 \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f'(x)}{g'(x)}\). 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

\(\lim_{x
ightarrow1}\frac{2x - 1}{\frac{1}{x}}=\lim_{x
ightarrow1}(2x - 1)x\).

Step4: Substitute \(x = 1\)

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

Answer:

1