Problem

Evaluate the limit \[ \lim _{t \rightarrow 0} \frac{\frac{1}{(6+t)^{2}}-\frac{1}{36}}{t} \]

Solution

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: \[\lim _{t \rightarrow 0} \frac{-12 - t}{36(6+t)^2} = \frac{-12}{36*6^2} = \boxed{-\frac{1}{18}}\]

From Solvely APP
Source: https://solvelyapp.com/problems/39998/

Get free Solvely APP to solve your own problems!

solvely Solvely
Download