Checking for discontinuity of $f(x)= sqrt{(1+p^2)/(1+q^2)}$ if $x = p/q neq 0$, $f(x) = x$ otherwise.
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Let $f$ maps from $mathbb{R}$ to $mathbb{R}$. Be given by
$$f(x)=
begin{cases}
sqrt{(1+p^2)/(1+q^2)},& x=p/q neq 0,\
x, & text{otherwise}.
end{cases}
$$
Then
a) $f$ is continuous on $big((0,infty) cap mathbb{Q'}big) cup {0,1}$. (Note that $mathbb{Q}'$ denote $mathbb{Q}$ complement).
b) If $a in (-infty,0)$ then $lim_{xto a} f(x)$ does not exists.
c) If $a in [0,infty)$, then $lim_{xto a} f(x)=a$
I know it is discontinuous, and in the answer sheet a, b, c are correct options. But I confused about option (b), how limit doesn't exist for negative number and also confused about option (a), how it is continuous on 0. I have tried a lot, eventually leads to headache, please help me, I shall be very thankful for it. Thanks in advance.
calculus continuity
edited Nov 20 at 5:30
Brahadeesh
5,98542159
asked Nov 18 at 18:20
user532616
214
add a comment |
up vote
-2
down vote
favorite
Let $f$ maps from $mathbb{R}$ to $mathbb{R}$. Be given by
$$f(x)=
begin{cases}
sqrt{(1+p^2)/(1+q^2)},& x=p/q neq 0,\
x, & text{otherwise}.
end{cases}
$$
Then
a) $f$ is continuous on $big((0,infty) cap mathbb{Q'}big) cup {0,1}$. (Note that $mathbb{Q}'$ denote $mathbb{Q}$ complement).
b) If $a in (-infty,0)$ then $lim_{xto a} f(x)$ does not exists.
c) If $a in [0,infty)$, then $lim_{xto a} f(x)=a$
I know it is discontinuous, and in the answer sheet a, b, c are correct options. But I confused about option (b), how limit doesn't exist for negative number and also confused about option (a), how it is continuous on 0. I have tried a lot, eventually leads to headache, please help me, I shall be very thankful for it. Thanks in advance.
calculus continuity
edited Nov 20 at 5:30
Brahadeesh
5,98542159
asked Nov 18 at 18:20
user532616
214
I have edited the formatting, let me know if the question is still accurate. Also, when you define $f(x)$ for $x = p/q$, are you assuming that $p/q$ is a rational with no common factors between $p$ and $q$?
– Brahadeesh
Nov 20 at 5:32
No such thing is mentioned, this question belongs to a test series of csir net
– user532616
Nov 20 at 15:47
Hmm, that’s odd...
– Brahadeesh
Nov 20 at 16:08
Option (b) is true for this reason: suppose $a < 0$ and $x to a$. If $x$ is rational, then $f(x)$ is positive and if $x$ is irrational then $f(x)$ is negative. So, the limit cannot exist (you need to complete the argument, but it should be easy from here).
– Brahadeesh
Nov 21 at 6:01
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let $f$ maps from $mathbb{R}$ to $mathbb{R}$. Be given by
$$f(x)=
begin{cases}
sqrt{(1+p^2)/(1+q^2)},& x=p/q neq 0,\
x, & text{otherwise}.
end{cases}
$$
Then
a) $f$ is continuous on $big((0,infty) cap mathbb{Q'}big) cup {0,1}$. (Note that $mathbb{Q}'$ denote $mathbb{Q}$ complement).
b) If $a in (-infty,0)$ then $lim_{xto a} f(x)$ does not exists.
c) If $a in [0,infty)$, then $lim_{xto a} f(x)=a$
I know it is discontinuous, and in the answer sheet a, b, c are correct options. But I confused about option (b), how limit doesn't exist for negative number and also confused about option (a), how it is continuous on 0. I have tried a lot, eventually leads to headache, please help me, I shall be very thankful for it. Thanks in advance.
calculus continuity
edited Nov 20 at 5:30
Brahadeesh
5,98542159
asked Nov 18 at 18:20
user532616
214
Let $f$ maps from $mathbb{R}$ to $mathbb{R}$. Be given by
$$f(x)=
begin{cases}
sqrt{(1+p^2)/(1+q^2)},& x=p/q neq 0,\
x, & text{otherwise}.
end{cases}
$$
Then
a) $f$ is continuous on $big((0,infty) cap mathbb{Q'}big) cup {0,1}$. (Note that $mathbb{Q}'$ denote $mathbb{Q}$ complement).
b) If $a in (-infty,0)$ then $lim_{xto a} f(x)$ does not exists.
c) If $a in [0,infty)$, then $lim_{xto a} f(x)=a$
I know it is discontinuous, and in the answer sheet a, b, c are correct options. But I confused about option (b), how limit doesn't exist for negative number and also confused about option (a), how it is continuous on 0. I have tried a lot, eventually leads to headache, please help me, I shall be very thankful for it. Thanks in advance.
calculus continuity
calculus continuity
edited Nov 20 at 5:30
Brahadeesh
5,98542159
asked Nov 18 at 18:20
user532616
214
edited Nov 20 at 5:30
Brahadeesh
5,98542159
asked Nov 18 at 18:20
user532616
214
edited Nov 20 at 5:30
Brahadeesh
5,98542159
edited Nov 20 at 5:30
Brahadeesh
5,98542159
edited Nov 20 at 5:30
Brahadeesh
5,98542159
5,98542159
asked Nov 18 at 18:20
user532616
214
asked Nov 18 at 18:20
user532616
214
asked Nov 18 at 18:20
user532616
214
214
I have edited the formatting, let me know if the question is still accurate. Also, when you define $f(x)$ for $x = p/q$, are you assuming that $p/q$ is a rational with no common factors between $p$ and $q$?
– Brahadeesh
Nov 20 at 5:32
No such thing is mentioned, this question belongs to a test series of csir net
– user532616
Nov 20 at 15:47
Hmm, that’s odd...
– Brahadeesh
Nov 20 at 16:08
Option (b) is true for this reason: suppose $a < 0$ and $x to a$. If $x$ is rational, then $f(x)$ is positive and if $x$ is irrational then $f(x)$ is negative. So, the limit cannot exist (you need to complete the argument, but it should be easy from here).
– Brahadeesh
Nov 21 at 6:01
add a comment |
I have edited the formatting, let me know if the question is still accurate. Also, when you define $f(x)$ for $x = p/q$, are you assuming that $p/q$ is a rational with no common factors between $p$ and $q$?
– Brahadeesh
Nov 20 at 5:32
No such thing is mentioned, this question belongs to a test series of csir net
– user532616
Nov 20 at 15:47
Hmm, that’s odd...
– Brahadeesh
Nov 20 at 16:08
Option (b) is true for this reason: suppose $a < 0$ and $x to a$. If $x$ is rational, then $f(x)$ is positive and if $x$ is irrational then $f(x)$ is negative. So, the limit cannot exist (you need to complete the argument, but it should be easy from here).
– Brahadeesh
Nov 21 at 6:01
I have edited the formatting, let me know if the question is still accurate. Also, when you define $f(x)$ for $x = p/q$, are you assuming that $p/q$ is a rational with no common factors between $p$ and $q$?
– Brahadeesh
Nov 20 at 5:32
I have edited the formatting, let me know if the question is still accurate. Also, when you define $f(x)$ for $x = p/q$, are you assuming that $p/q$ is a rational with no common factors between $p$ and $q$?
– Brahadeesh
Nov 20 at 5:32
No such thing is mentioned, this question belongs to a test series of csir net
– user532616
Nov 20 at 15:47
No such thing is mentioned, this question belongs to a test series of csir net
– user532616
Nov 20 at 15:47
Hmm, that’s odd...
– Brahadeesh
Nov 20 at 16:08
Hmm, that’s odd...
– Brahadeesh
Nov 20 at 16:08
Option (b) is true for this reason: suppose $a < 0$ and $x to a$. If $x$ is rational, then $f(x)$ is positive and if $x$ is irrational then $f(x)$ is negative. So, the limit cannot exist (you need to complete the argument, but it should be easy from here).
– Brahadeesh
Nov 21 at 6:01
Option (b) is true for this reason: suppose $a < 0$ and $x to a$. If $x$ is rational, then $f(x)$ is positive and if $x$ is irrational then $f(x)$ is negative. So, the limit cannot exist (you need to complete the argument, but it should be easy from here).
– Brahadeesh
Nov 21 at 6:01
add a comment |
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I have edited the formatting, let me know if the question is still accurate. Also, when you define $f(x)$ for $x = p/q$, are you assuming that $p/q$ is a rational with no common factors between $p$ and $q$?
– Brahadeesh
Nov 20 at 5:32
No such thing is mentioned, this question belongs to a test series of csir net
– user532616
Nov 20 at 15:47
Hmm, that’s odd...
– Brahadeesh
Nov 20 at 16:08
Option (b) is true for this reason: suppose $a < 0$ and $x to a$. If $x$ is rational, then $f(x)$ is positive and if $x$ is irrational then $f(x)$ is negative. So, the limit cannot exist (you need to complete the argument, but it should be easy from here).
– Brahadeesh
Nov 21 at 6:01