How do you find the limit of ln(100-n^2) as n approaches 10 from the negative side? No calculator and L'Hopital's rule is not allowed

Question

lim ln(!00-n²⁾

x→10^-

 

I'm not allowed to use a calculator for this problem and we haven't been taught L'Hopital's Rule. Please help!

Answer 1

ln(100-n²) = ln [(10-n)(10+n)] = ln(10-n) + ln(10+n) = Z

 

lim as n→10^- of Z  = ln 0.0001+ ln 20  (using 0.9999 as surrogate for 10^-)

and so the answer is -infinity!