Answer
$\displaystyle\lim_{x\rightarrow 1^{+}} \dfrac{\sec^{-1} x}{\sqrt{x-1}}\approx 1.4142$
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{x\rightarrow 1^{+}} \dfrac{\sec^{-1} x}{\sqrt{x-1}}$
Compute $\dfrac{\sec^{-1} x}{\sqrt{x-1}}$ for values of $x$ close to 1, approaching from the right:
$\dfrac{\sec^{-1} 2}{\sqrt{2-1}}\approx 1.0471976$
$\dfrac{\sec^{-1} 1.5}{\sqrt{1.5-1}}\approx 1.1894507$
$\dfrac{\sec^{-1} 1.1}{\sqrt{1.1-1}}\approx 1.3588297$
$\dfrac{\sec^{-1} 1.01}{\sqrt{1.01-1}}\approx 1.4083587$
$\dfrac{\sec^{-1} 1.001}{\sqrt{1.001-1}}\approx 1.4136247$
$\dfrac{\sec^{-1} 1.0001}{\sqrt{1.0001-1}}\approx 1.4141546$
$\dfrac{\sec^{-1} 1.00001}{\sqrt{1.00001-1}}\approx 1.4142077$
$\dfrac{\sec^{-1} 1.000001}{\sqrt{1.000001-1}}\approx 1.414213$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 1^{+}} \dfrac{\sec^{-1} x}{\sqrt{x-1}}\approx 1.4142$
