Question

use l'hopital's rule to find the limit.

lim ln(x^2 + 11x) / lnx
x->0

Answer 1

differentiate top and bottom

Answer 2

= limit [ (2x+11) / (x^2 +11x) ] / [1/x]

Answer 3

\[\lim_{x \rightarrow 0}= \frac{x (2x +11) }{x^2 +11x} \]

Answer 4

\[= \lim_{x \rightarrow 0} \frac{2x^2 +11x}{x^2 +11x}\]

Answer 5

still 0/0

differentiate again

Answer 6

\[\lim_{x \rightarrow 0}= \frac{4x + 11}{2x+11} \]

Answer 7

now, sub in x=0

you get 11/11 = 1

limit is 1