Mixing
Find all parameters $a,b,c,d in mathbb R$ for which the function $f: mathbb R rightarrow mathbb R$ a) has a...
$begingroup$
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
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
asked Jan 5 at 15:26
MP3129MP3129
2237
$endgroup$
|
show 3 more comments
$begingroup$
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
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
asked Jan 5 at 15:26
MP3129MP3129
2237
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
asked Jan 5 at 15:26
MP3129MP3129
2237
$endgroup$
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
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
asked Jan 5 at 15:26
MP3129MP3129
2237
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
asked Jan 5 at 15:26
MP3129MP3129
2237
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
edited Jan 5 at 15:30
Cameron Buie
85.1k771155
85.1k771155
asked Jan 5 at 15:26
MP3129MP3129
2237
asked Jan 5 at 15:26
MP3129MP3129
2237
asked Jan 5 at 15:26
MP3129MP3129
2237
2237
$begingroup$
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
1 Answer
1
active
oldest
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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 $b=frac{ln(2)ln(7)}2.$
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
$endgroup$
add a comment |
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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 $b=frac{ln(2)ln(7)}2.$
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
$endgroup$
add a comment |
$begingroup$
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 $b=frac{ln(2)ln(7)}2.$
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
$endgroup$
add a comment |
$begingroup$
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 $b=frac{ln(2)ln(7)}2.$
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
$endgroup$
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 $b=frac{ln(2)ln(7)}2.$
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 16:24
Cameron BuieCameron Buie
85.1k771155
85.1k771155
add a comment |
add a comment |
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$begingroup$
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
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@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
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– MP3129
Jan 5 at 15:41
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@CameronBuie I have not a Taylor's Theorem at my university and I mustn't use facts which I did not have a lecture
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– MP3129
Jan 5 at 15:43
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@MP3129: Okay. What about L'Hôpital's Rule?
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– Cameron Buie
Jan 5 at 15:43