Evaluate the limit
\[
\lim _{t \rightarrow 0} \frac{\frac{1}{(6+t)^{2}}-\frac{1}{36}}{t}
\]
Answer
Finally, we can take the limit as t approaches 0: \[\lim _{t \rightarrow 0} \frac{-12 - t}{36(6+t)^2} = \frac{-12}{36*6^2} = \boxed{-\frac{1}{18}}\]
Step 1 ：First, we can rewrite the expression inside the limit as follows: \[\frac{1}{(6+t)^{2}}-\frac{1}{36} = \frac{36 - (6+t)^2}{36(6+t)^2}\]
Step 2 ：Then, we can simplify the numerator: \[36 - (6+t)^2 = 36 - 36 - 12t - t^2 = -12t - t^2\]
Step 3 ：Substitute the simplified numerator back into the original expression: \[\frac{-12t - t^2}{36(6+t)^2}\]
Step 4 ：Next, we can divide the numerator and the denominator by t: \[\frac{-12 - t}{36(6+t)^2}\]
Step 5 ：Finally, we can take the limit as t approaches 0: \[\lim _{t \rightarrow 0} \frac{-12 - t}{36(6+t)^2} = \frac{-12}{36*6^2} = \boxed{-\frac{1}{18}}\]
