Answer
$$\infty $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{\sqrt y }} \cr
& {\text{Evaluate }} \cr
& \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{\sqrt y }} = \frac{{\mathop {\lim }\limits_{y \to {0^ + }} {{\ln }^{10}}y}}{{\mathop {\lim }\limits_{y \to {0^ + }} \sqrt y }} = \frac{\infty }{0} \cr
& \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{\sqrt y }} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{{y^{1/2}}}} \cr
& {\text{Use the property }}{x^a} = {e^{a\ln x}} \cr
& \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{{y^{1/2}}}} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{{e^{\frac{1}{2}\ln y}}}} \cr
& {\text{Apply }}\frac{1}{{{a^u}}} = {a^{ - u}},{\text{ then}} \cr
& \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{{\ln }^{10}}y}}{{{e^{\frac{1}{2}\ln y}}}} = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {{{\ln }^{10}}y \cdot {e^{ - \frac{1}{2}\ln y}}} \right] \cr
& {\text{Product property of limits}} \cr
& = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {{{\ln }^{10}}y} \right]\mathop {\lim }\limits_{y \to {0^ + }} \left[ {{e^{ - \frac{1}{2}\ln y}}} \right] \cr
& {\text{Evaluating}} \cr
& = \left[ {{{\ln }^{10}}\left( {{0^ + }} \right)} \right]\left[ {{e^{ - \frac{1}{2}\ln \left( {{0^ + }} \right)}}} \right] \cr
& = \left( \infty \right)\left[ {{e^{ - \frac{1}{2}\left( { - \infty } \right)}}} \right] \cr
& = \left( \infty \right)\left[ {{e^\infty }} \right] \cr
& = \infty \cr} $$
