Find all parameters $a,b,c,d in mathbb R$ for which the function $f: mathbb R rightarrow mathbb R$ a) has a...












1












$begingroup$


My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    $endgroup$
    – Marine Galantin
    Jan 5 at 15:30










  • $begingroup$
    @MP3129 Are you familiar with Taylor's Theorem?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:32










  • $begingroup$
    @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    $endgroup$
    – MP3129
    Jan 5 at 15:41












  • $begingroup$
    @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    $endgroup$
    – MP3129
    Jan 5 at 15:43






  • 1




    $begingroup$
    @MP3129: Okay. What about L'Hôpital's Rule?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:43
















1












$begingroup$


My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    $endgroup$
    – Marine Galantin
    Jan 5 at 15:30










  • $begingroup$
    @MP3129 Are you familiar with Taylor's Theorem?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:32










  • $begingroup$
    @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    $endgroup$
    – MP3129
    Jan 5 at 15:41












  • $begingroup$
    @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    $endgroup$
    – MP3129
    Jan 5 at 15:43






  • 1




    $begingroup$
    @MP3129: Okay. What about L'Hôpital's Rule?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:43














1












1








1


1



$begingroup$


My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis







share|cite|improve this question











$endgroup$




My function $f: mathbb R rightarrow mathbb R$ it is given a pattern:
$$f(x)=begin{cases} frac{2^{7^{x}-1}-a}{ln(1-x)},& x<0\ b,& x=0 \ frac{sinleft(c sqrt{x^{2}+d^{2}x}right) }{x},& x>0 end{cases}$$ for $$a,b,c,d in mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$



That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.



I thought about doing some substitution: for example $y=frac{1}{x}$, then for $x rightarrow 0^mp$ I have $y rightarrow pm infty$ and but I still can not solve this.



Can I count on any tips?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 15:30









Cameron Buie

86.4k773161




86.4k773161










asked Jan 5 at 15:26









MP3129MP3129

731211




731211












  • $begingroup$
    so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    $endgroup$
    – Marine Galantin
    Jan 5 at 15:30










  • $begingroup$
    @MP3129 Are you familiar with Taylor's Theorem?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:32










  • $begingroup$
    @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    $endgroup$
    – MP3129
    Jan 5 at 15:41












  • $begingroup$
    @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    $endgroup$
    – MP3129
    Jan 5 at 15:43






  • 1




    $begingroup$
    @MP3129: Okay. What about L'Hôpital's Rule?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:43


















  • $begingroup$
    so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
    $endgroup$
    – Marine Galantin
    Jan 5 at 15:30










  • $begingroup$
    @MP3129 Are you familiar with Taylor's Theorem?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:32










  • $begingroup$
    @MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
    $endgroup$
    – MP3129
    Jan 5 at 15:41












  • $begingroup$
    @CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
    $endgroup$
    – MP3129
    Jan 5 at 15:43






  • 1




    $begingroup$
    @MP3129: Okay. What about L'Hôpital's Rule?
    $endgroup$
    – Cameron Buie
    Jan 5 at 15:43
















$begingroup$
so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
$endgroup$
– Marine Galantin
Jan 5 at 15:30




$begingroup$
so the goal is to find the parameters such that the function defined for x negative, x positive and x = 0 is continuous ?
$endgroup$
– Marine Galantin
Jan 5 at 15:30












$begingroup$
@MP3129 Are you familiar with Taylor's Theorem?
$endgroup$
– Cameron Buie
Jan 5 at 15:32




$begingroup$
@MP3129 Are you familiar with Taylor's Theorem?
$endgroup$
– Cameron Buie
Jan 5 at 15:32












$begingroup$
@MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
$endgroup$
– MP3129
Jan 5 at 15:41






$begingroup$
@MarineGalantin in point (a) I need to find the parameters $ a, b, c, d in mathbb R $ for which the function has a limit in point 0 and in point (b) which is a separate case I have to indicate parameters for which the function is continuous in the whole field
$endgroup$
– MP3129
Jan 5 at 15:41














$begingroup$
@CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
$endgroup$
– MP3129
Jan 5 at 15:43




$begingroup$
@CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
$endgroup$
– MP3129
Jan 5 at 15:43




1




1




$begingroup$
@MP3129: Okay. What about L'Hôpital's Rule?
$endgroup$
– Cameron Buie
Jan 5 at 15:43




$begingroup$
@MP3129: Okay. What about L'Hôpital's Rule?
$endgroup$
– Cameron Buie
Jan 5 at 15:43










1 Answer
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$begingroup$

The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






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    $begingroup$

    The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



    If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



    Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



      If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



      Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



        If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



        Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$






        share|cite|improve this answer









        $endgroup$



        The first thing to note is that as $xnearrow 0,$ the denominator $ln(1-x)$ vanishes, so the only way to avoid $bigl|f(x)bigr|to+infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-ato2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.



        If you're familiar with the result that $$lim_{tto 0}frac{sin(t)}t=1,tag{1}$$ then we can tackle the right-hand limit as follows. Since $lim_{xsearrow0}csqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$lim_{xsearrow0}frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}=1,$$ so, since $$frac{sinleft(csqrt{x^2+d^2x}right)}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdotfrac{csqrt{x^2+d^2x}}{x}=frac{sinleft(csqrt{x^2+d^2x}right)}{csqrt{x^2+d^2x}}cdot csqrt{1+frac{d^2}x}$$ whenever $x>0,$ then $lim_{xsearrow0}f(x)=+infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=lim_{xnearrow0}f(x),$ assuming the left-hand limit exists.



        Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$begin{eqnarray}lim_{xnearrow0}frac{2^{7^x-1}-1}{ln(1-x)} &=& lim_{xnearrow0}cfrac{ln(2)ln(7)7^{x}2^{7x-1}}{-frac1{1-x}}\ &=& lim_{xnearrow0}(x-1)ln(2)ln(7)7^{x}2^{7x-1}\ &=& (0-1)ln(2)ln(7)7^{0}2^{0-1}\ &=& -frac{ln(2)ln(7)}2,end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=frac{ln(2)ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=frac{ln(2)ln(7)}2.$







        share|cite|improve this answer












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        answered Jan 5 at 16:24









        Cameron BuieCameron Buie

        86.4k773161




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            Probability when a professor distributes a quiz and homework assignment to a class of n students.

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            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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            Aardman Animations

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            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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            Are they similar matrix

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            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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