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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            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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            “minimization” problem in Euclidean space related to orthonormal basis

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            3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
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