Question

limit as y approaches infinity of 1/(yln1).i rewrote it to be (1/y)/lny, but it still doesn't even after applying the rule.

Answer 1

you get negative sign after applying L'Hopitals, hence -infty.

Answer 2

don't u get -1/x ?

Answer 3

\[\lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{\ln x}=\lim_{x \rightarrow 0^+} \frac{-\frac{1}{x^2}}{\frac{1}{x}}=\lim_{x \rightarrow 0^+}-\frac{1}{x}=-\infty\]

Answer 4

got it. it was a algebraic mistake on my part thanks hartnn and eseidl

Answer 5

welcome

Answer 6

hmm.. welcome ^_^