Question:

The value of the limit \(\lim_{t \to 0} \frac{(5-t)^2 - 25}{t}\) is equal to:

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Always look to simplify the expression first in limit problems, which often allows for straightforward evaluation without needing L'Hôpital's rule or more complex methods.
Updated On: Mar 12, 2025
  • -10
  • -5
  • 10
  • 5
  • 0
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The Correct Option is A

Solution and Explanation

First, expand and simplify the expression within the limit: \[ (5-t)^2 - 25 = (25 - 10t + t^2) - 25 = -10t + t^2 \] The limit becomes: \[ \lim_{t \to 0} \frac{-10t + t^2}{t} \] Simplify the expression by cancelling \(t\) from the numerator and the denominator: \[ \lim_{t \to 0} (-10 + t) \] As \(t\) approaches 0, the limit of the expression is: \[ -10 + 0 = -10 \] Therefore, the value of the limit is \(-10\).
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