Question

limits help please

Answer 1

\[\lim_{x \rightarrow 1^-} \frac{ 3x }{ x-1 }\]

Answer 2

\[=\lim_{x \rightarrow 1-}\frac{ -3x }{ 1-x }=\lim_{x \rightarrow 1-}\left( -3x \right)\left( 1-x \right)^{-1}\]
\[=-3\lim_{x \rightarrow 1-}x \left( 1+x+x^2+x^3+.... \right)\rightarrow - \infty \]
so limit does not exist.

Answer 3

Thanks!

Answer 4

yw