Check my proof that right-sided limit doesn't exist here
Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.
My solution : Since $f$ is continuous, we have
$lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $
However, the last limit doesn't exist and hence the result.
real-analysis limits
asked Nov 25 at 19:58
user69503
626
add a comment |
Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.
My solution : Since $f$ is continuous, we have
$lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $
However, the last limit doesn't exist and hence the result.
real-analysis limits
asked Nov 25 at 19:58
user69503
626
add a comment |
Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.
My solution : Since $f$ is continuous, we have
$lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $
However, the last limit doesn't exist and hence the result.
real-analysis limits
asked Nov 25 at 19:58
user69503
626
Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.
My solution : Since $f$ is continuous, we have
$lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $
However, the last limit doesn't exist and hence the result.
real-analysis limits
real-analysis limits
asked Nov 25 at 19:58
user69503
626
asked Nov 25 at 19:58
user69503
626
asked Nov 25 at 19:58
user69503
626
asked Nov 25 at 19:58
user69503
626
asked Nov 25 at 19:58
user69503
626
626
add a comment |
add a comment |
1 Answer
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We have that
$$frac{1+x}{sqrt x} to infty$$
then the limit doesn't exist, to prove that we can consider
$x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$
$x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$
answered Nov 25 at 20:04
gimusi
1
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
1
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
1
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
|
show 1 more comment
Your Answer
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1 Answer
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We have that
$$frac{1+x}{sqrt x} to infty$$
then the limit doesn't exist, to prove that we can consider
$x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$
$x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$
answered Nov 25 at 20:04
gimusi
1
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
1
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
1
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
|
show 1 more comment
We have that
$$frac{1+x}{sqrt x} to infty$$
then the limit doesn't exist, to prove that we can consider
$x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$
$x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$
answered Nov 25 at 20:04
gimusi
1
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
1
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
1
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
|
show 1 more comment
We have that
$$frac{1+x}{sqrt x} to infty$$
then the limit doesn't exist, to prove that we can consider
$x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$
$x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$
answered Nov 25 at 20:04
gimusi
1
We have that
$$frac{1+x}{sqrt x} to infty$$
then the limit doesn't exist, to prove that we can consider
$x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$
$x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$
answered Nov 25 at 20:04
gimusi
1
edited Nov 25 at 20:05
answered Nov 25 at 20:04
gimusi
1
answered Nov 25 at 20:04
gimusi
1
answered Nov 25 at 20:04
gimusi
1
1
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
1
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
1
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
|
show 1 more comment
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
1
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
1
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
Thank you! What about my proof? Is it all right or does it have any flaws?
– user69503
Nov 25 at 20:05
1
1
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
@user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
– gimusi
Nov 25 at 20:06
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
– user69503
Nov 25 at 20:14
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
@user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
– gimusi
Nov 25 at 20:19
1
1
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
@user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
– gimusi
Nov 25 at 20:27
|
show 1 more comment
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