# 355 by tanjawi1993

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‫ﺣﺴﺎب اﻟﻨﻬﺎﻳﺎت‬

‫- ﻟﺪﻳﻨﺎ :‬
‫1 + 3‪x‬‬
‫‪lim f ( x) = lim‬‬
‫∞−→ ‪x‬‬        ‫∞−→ ‪x‬‬     ‫‪x‬‬

‫3‪x‬‬
‫‪= lim‬‬
‫∞−→ ‪x‬‬   ‫‪x‬‬
‫∞+ = ‪= lim x‬‬
‫2‬
‫∞−→ ‪x‬‬

‫ﻟﺪﻳﻨﺎ‬
‫1+ ‪x + x − 2x‬‬
‫4‬       ‫3‬   ‫2‬
‫‪lim f ( x) = lim‬‬
‫∞+→ ‪x‬‬        ‫∞+→ ‪x‬‬         ‫3‪x‬‬
‫4‪x‬‬
‫‪= lim‬‬
‫∞+→ ‪x‬‬   ‫3‪x‬‬
‫∞+ = ‪= lim x‬‬
‫∞+→ ‪x‬‬

‫1 + 3‪x‬‬
‫‪lim f ( x) = lim‬‬          ‫ﻟﺪﻳﻨﺎ : ∞− =‬
‫−‬
‫0→ ‪x‬‬        ‫0→ ‪x‬‬‫−‬
‫3‪x‬‬
‫ﻷن : 1 = )1 + 3 ‪ lim ( x‬و −0 = ‪lim −x‬‬
‫−‬
‫0→ ‪x‬‬               ‫0→ ‪x‬‬
‫ﻟﺪﻳﻨﺎ :‬
‫1+ ‪x + x − 2x‬‬
‫4‬      ‫3‬           ‫2‬
‫‪lim f ( x) = lim‬‬             ‫∞+ =‬
‫0→ ‪x‬‬ ‫+‬
‫0→ ‪x‬‬     ‫+‬
‫3‪x‬‬
‫ﻷن 1 + 2 ‪ lim x 4 + x3 − 2 x‬و +0 = 3‪lim x‬‬
‫+‬               ‫+‬
‫0→ ‪x‬‬                ‫0→ ‪x‬‬

‫2 أ ﻟﻨﺒﻴﻦ أن ) ‪ ( Δ‬ﻣﻘﺎرب ﻟﻠﻤﻨﺤﻨﻰ ﺑﺠﻮار ∞+‬
‫ﻟﺪﻳﻨﺎ :‬
‫⎛‬       ‫⎞ 1 2‬
‫)1 + ‪lim f ( x) − ( x + 1) = lim ⎜ x + 1 − + 3 ⎟ − ( x‬‬
‫∞+→ ‪x‬‬                   ‫∞+→ ‪x‬‬
‫⎝‬       ‫⎠ ‪x x‬‬
‫⎞ 1 2 ⎛‬
‫0 = ⎟ 3 + − ⎜ ‪lim‬‬
‫∞+→ ‪x‬‬
‫⎠ ‪⎝ x x‬‬

‫إذن اﻟﻤﺴﺘﻘﻴﻢ ) ‪ ( Δ‬ذو اﻟﻤﻌﺎدﻟﺔ 1 + ‪ y = x‬هﻮ ﺑﺎﻟﻔﻌﻞ ﻣﻘﺎرب ﻣﺎﺋﻞ ﻟﻠﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﺑﺠﻮار ∞+‬
‫ب – دراﺳﺔ وﺿﻊ ) ‪ (ζ‬ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ‪ ( Δ‬ﻋﻠﻰ اﻟﻤﺠﺎل [∞ + '1] ﻟﻜﻞ ‪ x‬ﻣﻦ [∞ + '1] ﻟﺪﻳﻨﺎ :‬
‫1 2−‬
‫= )1 + ‪f ( x ) − ( x‬‬   ‫+‬
‫3‪x x‬‬
‫2 ‪1 − 2x‬‬
‫=‬
‫3‪x‬‬
‫ﻟﻜﻞ ‪ x‬ﻣﻦ [∞+ ,1] ﻓﺈن 0 ≺ )1 + ‪( ∀x ∈ ]1 + ∞[ ) f ( x ) − ( x‬‬
‫'‬                                                      ‫3‪x‬‬      ‫وﺑﻤﺎ أن 0 ≺ 2 ‪ 1 − 2 x‬و 0‬
‫)‪(Δ‬‬   ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈﻧﻪ ﻓﻲ اﻟﻤﺠﺎل [∞+ ,1] ﻳﻜﻮن اﻟﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﺗﺤﺖ ﻣﻘﺎرﺑﻪ اﻟﻤﺎﺋﻞ‬
‫ج - ﺑﺎﻗﻲ اﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ )‪( C‬‬
‫ﻟﺪﻳﻨﺎ‬
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‫‪lim‬‬     ‫∞+ = ) ‪f ( x‬‬
‫∞−→ ‪x‬‬

‫)‪f ( x‬‬          ‫1 + 3‪x‬‬
‫‪lim‬‬              ‫‪= lim‬‬           ‫و‬
‫∞−→‪x‬‬       ‫‪x‬‬       ‫∞−→‪x‬‬     ‫2‪x‬‬
‫3‪x‬‬
‫‪= lim‬‬
‫∞−→ ‪x‬‬   ‫2‪x‬‬
‫‪= lim‬‬      ‫∞− = ‪x‬‬
‫∞−→‪x‬‬
‫إذن اﻟﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﻳﻘﺒﻞ ﺑﺠﻮار ∞− ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ إﺗﺠﺎهﻪ ﻣﺤﻮر اﻷراﺗﻴﺐ .‬
‫ﻟﺪﻳﻨﺎ‬
‫+‪= lim‬‬     ‫−‪ lim‬و ∞+ = ) ‪f ( x‬‬          ‫∞− = ) ‪f ( x‬‬
‫0→ ‪x‬‬                        ‫0→ ‪x‬‬

‫إذن اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ 0 = ‪ x‬هﻮ ﻣﻘﺎرب رأﺳﻲ ﻟﻠﻤﻨﺤﻨﻰ ) ‪(ζ‬‬
‫3 ﺗﺤﺪﻳﺪ ' ‪f‬‬
‫اﻟﺪاﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ آﻞ ﻣﻦ اﻟﻤﺠﺎﻟﻴﻦ [∞+ ,0] و [0 ,∞−] وﻟﺪﻳﻨﺎ :‬
‫ﻟﻜﻞ ‪ x‬ﻣﻦ [∞ + '0] :‬

‫−1+ ‪(x ) = ⎛ x‬‬
‫⎞ 1 2‬
‫‪f‬‬   ‫'‬
‫⎜‬         ‫+‬  ‫⎟‬
‫⎝‬        ‫⎠3‪x x‬‬
‫3 2‬
‫4 − 2 +1 =‬
‫‪x‬‬    ‫‪x‬‬
‫3 − 2‪x + 2 x‬‬
‫4‬
‫=‬
‫4‪x‬‬
‫4 − )1 + ‪x 4 + 2 x 2 + 1 − 4 ( x‬‬
‫2‬      ‫2‬

‫=‬                      ‫=‬
‫4‪x‬‬                   ‫4‪x‬‬

‫=‬
‫)3 + 2 ‪( x2 + 1 − 2 )( x2 + 1 + 2 ) = ( x2 − 1)( x‬‬
‫4‪x‬‬                        ‫4‪x‬‬
‫ﻟﻜﻞ ‪ x‬ﻣﻦ [0 ' ∞−] :‬
‫⎛‬      ‫⎞1‬
‫⎟ + 2‪f ' ( x ) = ⎜ x‬‬
‫⎝‬      ‫⎠‪x‬‬
‫1‬
‫2 − ‪= 2x‬‬
‫‪x‬‬
‫1 − ‪2x‬‬
‫3‬
‫=‬
‫2‪x‬‬
‫4 – ﺟﺪول ﺗﻐﻴﺮات ‪f‬‬
‫* ﻋﻠﻰ اﻟﻤﺠﺎل [∞ + '0] إﺷﺎرة ) ‪ f ( x‬هﻲ إﺷﺎرة 1 − ‪x‬‬
‫2‬                 ‫'‬

‫* ﻋﻠﻰ اﻟﻤﺠﺎل [0 ' ∞−] إﺷﺎرة ) ‪ f ' ( x‬هﻲ إﺷﺎرة 1 − 3 ‪2 x‬‬
‫ﻓﺈن 0 ≺ 3‪ 2 x‬وﻣﻨﻪ 0 ≺ 1 − 3 ‪ 2 x‬ﻟﻜﻞ ‪ x‬ﻣﻦ [0 '∞−]‬      ‫وﺑﻤﺎ أن 0 ≺ ‪x‬‬
‫إذن ﺟﺪول ﺗﻐﻴﺮات ‪ f‬هﻮ :‬
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‫) ‪(ζ‬‬   ‫5 – أ ﻧﻘﻄﺘﺎ اﻧﻌﻄﺎف اﻟﻤﻨﺤﻨﻰ‬
‫)1 + ‪2 ( x + 1) ( x 2 − x‬‬
‫= ) ‪( ∀x ∈ ]−∞, 0[ ) f "( x‬‬          ‫3‪x‬‬
‫* ﻟﺪﻳﻨﺎ‬

‫)1 + ‪− x − 1 2 ( x − x‬‬
‫2‬

‫=‬        ‫.‬
‫‪−x‬‬           ‫2‪x‬‬
‫0 1 + ‪ x 2 − x‬ﻟﻜﻞ ‪ x‬ﻣﻦ ) 0 ≺ ‪( Δ‬‬                               ‫وﺑﻤﺎ أن‬
‫ﻓﻲ اﻟﻤﺠﺎل [0 '∞−] هﻲ إﺷﺎرة )1 − ‪( − x‬‬              ‫ﻓﺈن إﺷﺎرة ) ‪f ' ( x‬‬
‫) 3 − 2 ‪−4 ( x‬‬
‫= ) ‪( ∀x ∈ ]0, +∞[ ) f ( x‬‬
‫''‬

‫5‪x‬‬
‫ﻟﺪﻳﻨﺎ :‬

‫=‬
‫4‬   ‫(‬         ‫()‬
‫3 +‪3−x x‬‬         ‫)‬
‫5‪x‬‬
‫(‬           ‫)‬
‫إذن إﺷﺎرة ) ‪ f ' ( x‬ﻓﻲ [∞ + '0] هﻲ إﺷﺎرة ‪3 − x‬‬
‫'‬

‫‪: f‬‬   ‫''‬
‫) ‪(x‬‬   ‫* اﻟﺠﺪول اﻟﺘﺎﻟﻲ ﻳﻌﻄﻲ إﺷﺎرة‬

‫‪ f‬ﺗﻨﻌﺪم ﻣﻊ ﺗﻐﻴﻴﺮ اﻹﺷﺎرة ﻓﻲ آﻞ ﻣﻦ )1− ( و 3‬           ‫''‬
‫اﻟﺪاﻟﺔ‬
‫إذن ﻟﻠﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﻧﻘﻄﺘﺎ اﻧﻌﻄﺎف هﻤﺎ ) 01− ( ‪ A‬هﻮ‬
‫'‬

‫⎛‬     ‫⎞ 3 3+4‬
‫,3 ⎜ ‪B‬‬
‫⎜‬             ‫⎟‬
‫⎟‬
‫⎝‬       ‫⎠ 3 3‬
‫ب * ﻣﻌﺎدﻟﺔ ﻟﻤﻤﺎس ) ‪ (ζ‬ﻓﻲ ) 01− ( ‪A‬‬
‫'‬

‫3 – ‪y = -3x‬‬
‫⎛‬      ‫⎞ 3 3+4‬
‫⎜‪B‬‬
‫⎜‬   ‫,3‬       ‫ﻣﻌﺎدﻟﺔ ﻟﻤﻤﺎس ) ‪ (ζ‬ﻓﻲ ⎟‬
‫⎝‬       ‫⎟ 3 3‬‫⎠‬
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4    3 3 −8
y=     x+
3     3 3
(ζ )   ‫6 – رﺳﻢ اﻟﻤﻨﺤﻨﻰ‬

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