Evaluate the limit
$lim_{t \to 0} \frac{\frac{1}{(6 + t)^{2}} - \frac{1}{36}}{t}$

Answer

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Answer

Finally, we can take the limit as t approaches 0: $lim_{t \to 0} \frac{- 12 - t}{36 (6 + t)^{2}} = \frac{- 12}{36 * 6^{2}} = - \frac{1}{18}$

Steps

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: $36 - (6 + t)^{2} = 36 - 36 - 12 t - t^{2} = - 12 t - t^{2}$

Step 3 ：Substitute the simplified numerator back into the original expression: $\frac{- 12 t - t^{2}}{36 (6 + t)^{2}}$

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: $lim_{t \to 0} \frac{- 12 - t}{36 (6 + t)^{2}} = \frac{- 12}{36 * 6^{2}} = - \frac{1}{18}$

Evaluate the limitlimt→01(6+t)2−136t

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Evaluate the limit
$lim_{t \to 0} \frac{\frac{1}{(6 + t)^{2}} - \frac{1}{36}}{t}$