evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim(x→1) (ln x)/(21x…

evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim(x→1) (ln x)/(21x - x² - 20) use lhopitals rule to rewrite the given limit so that it is not an indeterminate form. lim(x→1) (ln x)/(21x - x² - 20)=lim(x→1) ( ) evaluate the limit. lim(x→1) (ln x)/(21x - x² - 20)= (type an exact answer.)

evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim(x→1) (ln x)/(21x - x² - 20) use lhopitals rule to rewrite the given limit so that it is not an indeterminate form. lim(x→1) (ln x)/(21x - x² - 20)=lim(x→1) ( ) evaluate the limit. lim(x→1) (ln x)/(21x - x² - 20)= (type an exact answer.)

Answer

Explanation:

Step1: Check indeterminate form

When (x = 1), (\ln(1)=0) and (21\times1 - 1^{2}-20=21 - 1 - 20 = 0). So it's a (\frac{0}{0}) form and L'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of (y = \ln x) is (y'=\frac{1}{x}), and the derivative of (y=21x - x^{2}-20) is (y'=21 - 2x). So (\lim_{x\rightarrow1}\frac{\ln x}{21x - x^{2}-20}=\lim_{x\rightarrow1}\frac{\frac{1}{x}}{21 - 2x}).

Step3: Evaluate the new - limit

Substitute (x = 1) into (\frac{\frac{1}{x}}{21 - 2x}), we get (\frac{\frac{1}{1}}{21-2\times1}=\frac{1}{19}).

Answer:

(\frac{1}{19})