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- Politecnico di Milano
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- Analisi Matematica I e Geometria
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Limiti
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Andrea Landella, POLITECNICO DI MILANO 1 Risolvere i seguenti limiti usando quando possibile i limiti notevoli: 1. lim 𝑥→−2 𝑥2 − 𝑥 − 6 𝑥3 + 5𝑥2 + 8𝑥 + 4 = lim 𝑥→−2 (𝑥 − 3)(𝑥 + 2) (𝑥 + 1)(𝑥 + 2)2 = lim 𝑥→−2 (𝑥 − 3) (𝑥 + 1)(𝑥 + 2) = −∞ 2. lim 𝑥→±∞ 𝑥3 + 1 𝑥 − 1 = lim 𝑥→±∞ 𝑥3 + 𝑜(𝑥) 𝑥 + 𝑜(𝑥) = 𝑥2 = +∞ 3. lim 𝑥→+∞ √𝑥(√𝑥 − √1 + 𝑥) = lim 𝑥→+∞ − √𝑥 √𝑥 + √1 + 𝑥 = − lim 𝑥→+∞ √𝑥 2√𝑥 + 𝑜(𝑥) = − 1 2 + 4. lim 𝑥→+∞ √𝑥2 + 1 𝑥1/3 = lim 𝑥→+∞ 𝑥 + 𝑜(𝑥) 𝑥1/3 = +∞ 5. lim 𝑥→1 𝑥 + 1 − 2√𝑥 (𝑥 − 1)2 = lim 𝑥→1 (√𝑥 − 1) 2 (𝑥 − 1)2 = lim 𝑥→1 (√𝑥 − 1) 2 (√𝑥 − 1) 2 (√𝑥 + 1) 2 = lim 𝑥→1 1 (√𝑥 + 1) 2 = 1 4 6. lim 𝑥→0 1 − cos2𝑥 sen2 3𝑥 = lim 𝑥→0 1 − cos2𝑥 4𝑥2 ⋅ 9𝑥2 sen2 3𝑥 ⋅ 4 9 = lim 𝑥→0 1 − cos2𝑥 (2𝑥)2 ⋅ (3𝑥)2 sen2 3𝑥 ⋅ 4 9 = 2 9 7. lim 𝑥→0+ √1 − cos𝑥 𝑥 = lim 𝑥→0+ (1 − cos𝑥 𝑥2 ) 1/2 = √2 2 8. lim 𝑥→0+(log𝑥 − logsen2𝑥) = lim 𝑥→0+ log( 𝑥 sen2𝑥) = log[ lim 𝑥→0+ ( 2𝑥 sen2𝑥 ⋅ 1 2)] = −log2 9. lim 𝑥→+∞ log(√𝑥2 + 1 − 𝑥) = log( lim 𝑥→+∞ √𝑥2 + 1 − 𝑥 ⋅ √𝑥2 + 1 + 𝑥 √𝑥2 + 1 + 𝑥 ) = log lim 𝑥→+∞ 1 √𝑥2 + 1 + 𝑥 = −∞ 10. lim 𝑥→0 1 − cos𝑥 𝑒2𝑥 − 2𝑒𝑥 + 1 = lim 𝑥→0 1 − cos𝑥 𝑥2 ⋅ ( 𝑥 𝑒𝑥 − 1) 2 = 1 2 11. lim 𝑥→0+ 𝑒 log2 𝑥−2 log 𝑥−2 = exp lim 𝑥→0+ log2 𝑥 + 𝑜(𝑥) log𝑥 + 𝑜(𝑥) = lim 𝑥→0+𝑒log 𝑥 = lim 𝑥→0+𝑥 = 0+ 12. lim 𝑥→+∞ log(log𝑥 − 1 log𝑥 + 1) = log lim 𝑥→+∞ log𝑥 + 𝑜(𝑥) log𝑥 + 𝑜(𝑥) = log lim 𝑥→+∞ 1 = 0+ 13. lim 𝑥→0 1 − cos𝑥 log( 𝑥 𝑥 + 1) = lim 𝑥→0 1 − cos𝑥 𝑥2 ⋅ 𝑥2 log( 𝑥 𝑥 + 1) = lim 𝑥→0 𝑥2 2 log𝑥 − 2 log(𝑥 + 1) = 0− 14. lim 𝑥→0+ 1 − cos 𝑥4 sen2 𝑥 = lim 𝑥→0+ 1 − cos𝑥4 𝑥2 ⋅ 𝑥2 sen2 𝑥 = lim 𝑥→0+ 1 − cos𝑥4 𝑥2 ⋅ 𝑥2 𝑥2 = lim 𝑥→0+ 1 − cos𝑥4 𝑥4 ⋅ 𝑥2 = lim 𝑥→0+ 𝑥2 2 = 0+ 15. lim 𝑥→0 (sin2𝑥 𝑥 ) 𝑥+1 = lim 𝑥→0 (sin2𝑥 2𝑥 ⋅ 2) 𝑥+1 = lim 𝑥→0 2𝑥+1 = 2 16. lim 𝑥→+∞ √𝑥 √𝑥 + √𝑥 + √𝑥 = √ lim 𝑥→+∞ 𝑥 𝑥 + √𝑥 + √𝑥 = √ lim 𝑥→+∞ 𝑥 𝑥 + √𝑥 + 𝑜(𝑥) = √ lim 𝑥→+∞ 𝑥 𝑥 + 𝑜(𝑥) = 1− Andrea Landella, POLITECNICO DI MILANO 2 17. lim 𝑥→0+ 𝑥 1 log2 𝑥 = lim 𝑥→0 𝑥…
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