Exercici 1. 

a) Asímptota horitzontal de la funció: f left parenthesis x right parenthesis space equals fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction


Tal i com s'explica en l'apartat : Asímptotes horitzontals del recurs : Resum conceptes bàsics del  lliurament 1

              Cal estudiar aquests límits en l'infinit de la funció proposada:

space space stack lim space space space space space with x rightwards arrow negative infinity below f left parenthesis x right parenthesis equals space space bold space bold space stack l i m space space space space space space with x rightwards arrow negative infinity below bold space fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction bold equals fraction numerator bold plus bold infinity over denominator bold minus bold infinity end fraction bold space I n d e t e r m i n a c i ó comma space q u e space s apostrophe h a space d e space r e s o l d r e space u s a n t space l a space r e g l a space d e l s space g r a u s space stack lim space space space space space with straight x rightwards arrow negative infinity below space fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction space equals plus bold infinity bold space bold space bold italic p bold italic e bold italic r bold space bold italic q bold italic u bold italic è bold space bold italic e bold italic l bold space bold italic p bold italic o bold italic l bold italic i bold italic n bold italic o bold italic m bold italic i bold space bold italic d bold italic e bold space bold italic m bold italic a bold italic j bold italic o bold italic r bold space bold italic g bold italic r bold italic a bold italic u bold space bold italic é bold italic s bold space bold italic e bold italic l bold space bold italic d bold italic e bold italic l bold space bold italic n bold italic u bold italic m bold italic e bold italic r bold italic a bold italic d bold italic o bold italic r C o m space q u e space a q u e s t space l í m i t space bold italic n bold italic o bold space bold italic e bold italic s bold space bold italic u bold italic n bold space bold italic n bold italic o bold italic m bold italic b bold italic r bold italic e bold space bold italic r bold italic e bold italic a bold italic l space bold italic n bold italic o bold space bold italic h bold italic i bold space bold italic h bold italic a bold space bold italic a bold italic s bold italic í bold italic m bold italic p bold italic t bold italic o bold italic t bold italic a bold space bold italic h bold italic o bold italic r bold italic i bold italic t bold italic z bold italic o bold italic n bold italic t bold italic a bold italic l bold. space


space De space la space mateixa space manera space cal space calcular colon limit as space space space space space space space space straight x rightwards arrow plus infinity of straight f left parenthesis straight x right parenthesis equals space bold space bold space stack l i m space space space space space with x rightwards arrow plus infinity below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction bold equals fraction numerator bold plus bold infinity over denominator bold minus bold infinity end fraction bold space I n d e t e r m i n a c i ó comma space q u e space s apostrophe h a space d e space r e s o l d r e space u s a n t space l a space r e g l a space d e l s space g r a u s space stack l i m space space space space space with straight x rightwards arrow plus infinity below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction space equals space bold minus bold infinity bold space bold italic j bold italic a bold space bold italic q bold italic u bold italic e bold space bold italic e bold italic l bold space bold italic p bold italic o bold italic l bold italic i bold italic n bold italic o bold italic m bold italic i bold space bold italic d bold italic e bold space bold italic m bold italic a bold italic j bold italic o bold italic r bold space bold italic g bold italic r bold italic a bold italic u bold space bold italic é bold italic s bold space bold italic e bold italic l bold space bold italic d bold italic e bold italic l bold space bold italic n bold italic u bold italic m bold italic e bold italic r bold italic a bold italic d bold italic o bold italic r C o m space q u e space a q u e s t space l í m i t space bold italic n bold italic o bold space bold italic e bold italic s bold space bold italic u bold italic n bold space bold italic n bold italic o bold italic m bold italic b bold italic r bold italic e bold space bold italic r bold italic e bold italic a bold italic l space bold italic n bold italic o bold space bold italic h bold italic i bold space bold italic h bold italic a bold space bold italic a bold italic s bold italic í bold italic m bold italic p bold italic t bold italic o bold italic t bold italic a bold space bold italic h bold italic o bold italic r bold italic i bold italic t bold italic z bold italic o bold italic n bold italic t bold italic a bold italic l bold.



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Exercici 1.

b) Asímptota vertical de la funció: f left parenthesis x right parenthesis space equals fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction



Tal i com s'explica en l'apartat : Asímptotes verticals del recurs : Resum conceptes bàsics del  lliurament 1

Una funció té una asímptota vertical en x = a quan   limit as x rightwards arrow a of f left parenthesis x right parenthesis equals infinity

Majoritàriament, els possibles valors de x on pot passar això són els punts que no són del domini i els punts que sent del domini hi ha algun límit lateral que dóna   ∞

Un cop detectats aquests punts cal comprovar que  limit as x rightwards arrow a of f left parenthesis x right parenthesis equals infinity.


Per tant cal detectar el punts que no són del domini. En primer lloc busquem el domini.  Cal fer :   Denominador = 0

Resolem l'equació :

negative x plus 1 space equals 0 x equals 1 space space space A r a space c a l space c a l c u l a r space e l s space l í m i t s space l a t e r a l s space e n space x equals 1 limit as x rightwards arrow left parenthesis 1 right parenthesis to the power of plus of f left parenthesis x right parenthesis equals stack l i m with x rightwards arrow left parenthesis 1 right parenthesis to the power of plus below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction equals fraction numerator negative 1 over denominator 0 to the power of minus end fraction equals plus infinity stack l i m with x rightwards arrow left parenthesis 1 right parenthesis to the power of minus below f left parenthesis x right parenthesis equals stack l i m with x rightwards arrow left parenthesis 1 right parenthesis to the power of minus below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction equals fraction numerator negative 1 over denominator 0 to the power of plus end fraction equals negative infinity


Observeu que el  Dom(f(x))= straight real numbers minus left curly bracket 1 right curly bracket i que per tant hem estudiat els límits i hem comprovat que sí hi ha asímptota vertical en x =1


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Exercici 1.

c) Asímptota obliqua


Tal i com s'explica en el Resum dels conceptes bàsics del lliurament, cal fer

1.- m equals limit as x rightwards arrow infinity of fraction numerator f left parenthesis x right parenthesis over denominator x end fraction

    2.-  b equals space stack lim space space space with x rightwards arrow infinity below f left parenthesis x right parenthesis minus m x

    3.- Un cop hem trobat m i b l'asímptota obliqua serà la recta que té per equació    box enclose y equals m x plus b end enclose


                        Seguint el procediment explicat, calcularem

    limit as x rightwards arrow infinity of fraction numerator f left parenthesis x right parenthesis over denominator x end fraction equals limit as x rightwards arrow infinity of fraction numerator begin display style fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction end style over denominator x end fraction equals limit as x rightwards arrow infinity of space fraction numerator x squared minus 2 over denominator x left parenthesis negative x plus 1 right parenthesis end fraction equals limit as x rightwards arrow infinity of space fraction numerator x squared minus 2 over denominator negative x squared plus x end fraction equals fraction numerator 1 over denominator negative 1 end fraction equals negative 1 space left parenthesis e l s space g r a u s space d e l s space p o l i n o m i s space c o i n c i d e i x e n space i space h e m space d i v i d i t space e l s space c o e f i c i e n t s space d e l s space t e r m e s space d e space m a j o r space g r a u right parenthesis


    stack lim space space space with x rightwards arrow infinity below f left parenthesis x right parenthesis minus m x space equals stack lim space space space with x rightwards arrow infinity below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction minus left parenthesis negative 1 right parenthesis x equals stack lim space space space with x rightwards arrow infinity below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction plus 1 x equals stack lim space space space with x rightwards arrow infinity below fraction numerator x squared minus 2 over denominator negative x plus 1 end fraction plus fraction numerator x left parenthesis negative x plus 1 right parenthesis over denominator negative x plus 1 end fraction equals stack lim space space space with x rightwards arrow infinity below fraction numerator x ² minus 2 minus x ² plus x over denominator negative x plus 1 end fraction equals stack lim space space space with x rightwards arrow infinity below fraction numerator negative 2 plus x over denominator negative x plus 1 end fraction equals negative 1

                          Per tant l'asímptota obliqua és box enclose y equals m x plus b end enclose---->  box enclose y equals negative x minus 1 end enclose

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    Exercici 2. Calcula aquest límit, tot comprovant que és una indeterminació del tipus 1∞

    stack l i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction close parentheses to the power of negative x squared end exponent


    Resolució :

    Tal i com s'explica en l'apartat : anomenat "Resum conceptes bàsics del lliurament 1, apartat indeterminació del tipus 1∞

    Es pot resoldre seguint un procediment matemàtic o recordant la fórmula.

    Aquí ho farem usant la fórmula. 

    Primer comprovem que efectivament el límit proposat és del tipus 1∞

    stack l i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction close parentheses to the power of negative x squared end exponent equals open parentheses infinity over infinity close parentheses to the power of infinity L a space i n d e t e r m i n a c i ó space infinity over infinity space e s space p o t space r e s o l d r e space u s a n t space l a space r e g l a space d e l s space g r a u s space d e l space p o l i n o m i s

    stack lim space with x rightwards arrow infinity below space open parentheses fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction close parentheses equals limit as x rightwards arrow infinity of space open parentheses fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction close parentheses equals 1 over 1 equals 1  ja que els polinomis tenen el mateix grau.


    Per tant:


    stack l i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction close parentheses to the power of negative x squared end exponent equals open parentheses infinity over infinity close parentheses to the power of infinity equals 1 to the power of infinity



    Com el límit que ens proposen, una vegada substituït x per "infinit" obtenim la indeterminació bold 1 to the power of bold infinity  podem usar aquesta fórmula :

     box enclose stack l i m with x rightwards arrow infinity below f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent space space space equals space e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis f left parenthesis x right parenthesis negative 1 right parenthesis space times g left parenthesis x right parenthesis end exponent space end enclose


    box enclose stack l i m with x rightwards arrow infinity below f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent space space space equals space e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis f left parenthesis x right parenthesis negative 1 right parenthesis space times g left parenthesis x right parenthesis end exponent space equals space e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction negative 1 right parenthesis space times left parenthesis negative x squared right parenthesis end exponent end enclose o n space f left parenthesis x right parenthesis space equals fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction space space i space g left parenthesis x right parenthesis space equals negative x squared e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction negative 1 right parenthesis space times left parenthesis negative x squared right parenthesis end exponent equals e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis fraction numerator 3 x squared minus 1 over denominator 3 x squared plus 1 end fraction minus fraction numerator 3 x squared plus 1 over denominator 3 x squared plus 1 end fraction right parenthesis space times left parenthesis negative x squared right parenthesis end exponent equals e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis fraction numerator negative 2 over denominator 3 x squared plus 1 end fraction right parenthesis space times left parenthesis negative x squared right parenthesis end exponent equals equals e to the power of stack l i m space space with x rightwards arrow infinity below left parenthesis fraction numerator 2 x squared over denominator 3 x squared plus 1 end fraction right parenthesis space end exponent left parenthesis m i r a n t space g r a u s right parenthesis equals box enclose bold e to the power of bold 2 over bold 3 end exponent end enclose


    Exercici 3


    a) Cal calcular P(0) o sigui substituir t =0 en l'expressió algebraica de la funció.

    P left parenthesis 0 right parenthesis equals fraction numerator 0 squared plus 28 over denominator left parenthesis 0 plus 2 right parenthesis squared end fraction equals 28 over 4 equals 7 space m i l i o n s space d apostrophe h a b i tan t s


    b) A llarg termini:


    E s space d e m a n a space c a l c u l a r space a q u e s t space l í m i t colon limit as t rightwards arrow plus infinity of P left parenthesis t right parenthesis equals limit as t rightwards arrow plus infinity of fraction numerator t squared plus 28 over denominator left parenthesis t plus 2 right parenthesis squared end fraction equals limit as t rightwards arrow plus infinity of fraction numerator t squared plus 28 over denominator left parenthesis t squared plus 2 t plus 4 right parenthesis end fraction equals 1 space m i l i ó space d apostrophe h a b i tan t s P e r space c a l c u l a r space a q u e s t space l í m i t space h e m space m i r a t space e l s space g r a u s space d e l s space p o l i n o m i s.


    c) Per donar resposta a aquest apartat s'ha de calcular:



    E s space d e m a n a space c a l c u l a r space a q u e s t space l í m i t colon limit as t rightwards arrow negative 2 of P left parenthesis t right parenthesis equals limit as t rightwards arrow negative 2 of space fraction numerator t squared plus 28 over denominator left parenthesis t plus 2 right parenthesis squared end fraction equals limit as t rightwards arrow plus infinity of fraction numerator 4 plus 28 over denominator left parenthesis negative 2 plus 2 right parenthesis end fraction equals 32 over 0 equals infinity space m i l i o n s space d apostrophe h a b i tan t s
    Vol dir que 2 anys abans del moment actual la població era d'infinits habitants.
    Com això no té sentit, cal pensar que la fórmula que calcula la població no era aplicable en moments
    anteriors a t=0.
    He donat per bona qualsevol interpretació que tingui sentit







    Last modified: Tuesday, 15 October 2019, 9:50 AM