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Please help with calculus!?

Finding the limits, and my textbook is not really good. It gives answers, but does not explain how to find them. I think it is written graphing calculators, and I prefer to know how to resolve personally. Yes, could you explain how to solve this? 1) Limit as x approaches 2, (2 – x) / (x ^ 2 to 4) 2) s (t) =- 4.9t ^ 2 + 150 which gives the height in meters of an object that has dropped from a height of 150 meters. The velocity at time t = seconds is given by: limit as t approaches a, (s (a) – s (t)) / (a – t) Find the velocity of the object when t = 3 Thank you in advance for your help.

1) (2-x) / (x ^ 2 – 4) = (2-x) / [(x-2) (x +2)] = -1 / (x +2) The limit as x approaches 2 = -1 / (2 +2) = -1 / 4 ans: -1 / 4 2) s (t) =- 4.9t ^ 2 + 150 s (a) =- 4.9a ^ 2 + 150 s (a) – s (t) = (-4.9a ^ 2 + 150) – (-4.9t ^ 2 + 150) s (a) – s (t) = -4.9 (a ^ 2 – t ^ 2) s (a) – S (t) = -4.9 (a – t) (a + t) (s ( a) – s (t)) / (a) = -4.9 (a – t) (a + t) / (a) -4.9 = (A + t) ta as it approaches the speed -4.9 = (a + a) = -4.9 * 2A = -9.8 T Place = 3 or A = 3, = -9.8 * 3 = -29.4 speed m / s Answer: -29.4 M / s

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