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202 𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟏. 𝑎) 𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒 lim 𝑥→0 (𝑓(𝑥) + 𝑔(𝑥)) 2 = 4 𝑒 lim 𝑥→0 (𝑓(𝑥) − 𝑔(𝑥)) 2 = 2 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑒 lim 𝑥→0 (𝑓(𝑥)𝑔(𝑥)). 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 𝑜 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑞𝑢𝑜𝑐𝑖𝑒𝑛𝑡𝑒: (𝑓(𝑥) + 𝑔(𝑥)) 2 (𝑓(𝑥) − 𝑔(𝑥)) 2 = [𝑓(𝑥)]2 + 2. 𝑓(𝑥).𝑔(𝑥) + [𝑔(𝑥)]2 [𝑓(𝑥)]2 − 2. 𝑓(𝑥).𝑔(𝑥) + [𝑔(𝑥)]2 = 1 + 4𝑓(𝑥)𝑔(𝑥) (𝑓(𝑥) − 𝑔(𝑥)) 2 𝐶𝑜𝑚𝑜 lim 𝑥→0 (𝑓(𝑥)+ 𝑔(𝑥)) 2 𝑒 lim 𝑥→0 (𝑓(𝑥) − 𝑔(𝑥)) 2 𝑒𝑥𝑖𝑠𝑡𝑒𝑚,𝑐𝑜𝑚 𝑒𝑠𝑡𝑒 ú𝑙𝑡𝑖𝑚𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒 𝑑𝑒 𝑧𝑒𝑟𝑜,𝑒𝑛𝑡ã𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑜 𝑞𝑢𝑜𝑐𝑖𝑒𝑛𝑡𝑒 é 𝑜 𝑞𝑢𝑜𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑒, 𝑐𝑜𝑚𝑜 lim 𝑥→0 1 𝑒𝑥𝑖𝑠𝑡𝑒, 𝑐𝑜𝑛𝑐𝑙𝑢𝑖-𝑠𝑒 𝑝𝑒𝑙𝑜 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑚𝑒𝑚𝑏𝑟𝑜 𝑑𝑎 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑎𝑐𝑖𝑚𝑎 𝑞𝑢𝑒 lim 𝑥→0 (𝑓(𝑥)𝑔(𝑥)) 𝑡𝑎𝑚𝑏é𝑚 𝑒𝑥𝑖𝑠𝑡𝑒. 𝐿𝑜𝑔𝑜, lim 𝑥→0 (𝑓(𝑥) + 𝑔(𝑥)) 2 (𝑓(𝑥) − 𝑔(𝑥)) 2 = lim 𝑥→0 1 + lim 𝑥→0 4. 𝑓(𝑥)𝑔(𝑥) (𝑓(𝑥) − 𝑔(𝑥)) 2 lim 𝑥→0 (𝑓(𝑥) + 𝑔(𝑥)) 2 lim 𝑥→0 (𝑓(𝑥) − 𝑔(𝑥)) 2 = lim 𝑥→0 1 + lim 𝑥→0 (4.𝑓(𝑥)𝑔(𝑥)) lim 𝑥→0 (𝑓(𝑥) − 𝑔(𝑥)) 2 4 2 = 1+ 4 lim 𝑥→0 (𝑓(𝑥)𝑔(𝑥)) 2 2 = 1+ 2 lim 𝑥→0 (𝑓(𝑥)𝑔(𝑥)) lim 𝑥→0 (𝑓(𝑥)𝑔(𝑥)) = 1 2 𝑏) 𝐶𝑎𝑙𝑐𝑢𝑙𝑒 lim 𝑥→∞ (√𝑥 + √𝑥 + √𝑥 −√𝑥). lim 𝑥→∞ (√𝑥 + √𝑥 + √𝑥 −√𝑥) = lim 𝑥→∞ 𝑥 + √𝑥 + √𝑥 − 𝑥 √𝑥 +√𝑥 +√𝑥 + √𝑥 = lim 𝑥→∞ √𝑥 + √𝑥 √𝑥 +√𝑥 +√𝑥 + √𝑥