Answer
$\lim_{t\to4}\dfrac{\dfrac{1}{\sqrt{t}}-\dfrac{1}{2}}{t-4}=-\dfrac{1}{16}$
Work Step by Step
$\lim_{t\to4}\dfrac{\dfrac{1}{\sqrt{t}}-\dfrac{1}{2}}{t-4}$
Try to evaluate the limit applying direct substitution:
$\lim_{t\to4}\dfrac{\dfrac{1}{\sqrt{t}}-\dfrac{1}{2}}{t-4}=\dfrac{\dfrac{1}{\sqrt{4}}-\dfrac{1}{2}}{4-4}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Evaluate the subtraction in the numerator and simplify:
$\lim_{t\to4}\dfrac{\dfrac{1}{\sqrt{t}}-\dfrac{1}{2}}{t-4}=\lim_{t\to4}\dfrac{\dfrac{2-\sqrt{t}}{2\sqrt{t}}}{t-4}=\lim_{t\to4}\dfrac{2-\sqrt{t}}{2\sqrt{t}(t-4)}=...$
$...=\lim_{t\to4}\dfrac{-(\sqrt{t}-2)}{2\sqrt{t}(\sqrt{t}-2)(\sqrt{t}+2)}=\lim_{t\to4}\dfrac{-1}{2\sqrt{t}(\sqrt{t}+2)}=...$
Try to evaluate the limit using direct substitution again:
$\lim_{t\to4}\dfrac{-1}{2\sqrt{t}(\sqrt{t}+2)}=\dfrac{-1}{2\sqrt{4}(\sqrt{4}+2)}=\dfrac{-1}{(2)(2)(2+2)}=-\dfrac{1}{16}$
