Mixing
Prob. 15, Sec. 4.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: $lim_xto csqrtf(x)=sqrtlim_xto cf(x)$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Here is Prob. 15, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $A subseteq mathbbR$, let $f colon A to mathbbR$, and let $c in mathbbR$ be a cluster point of $A$. In addition, suppose that $f(x) geq 0$ for all $x in A$, and let $sqrtf$ be the function defined for $x in A$ by $left(sqrtfright)(x) = sqrtf(x)$. If $lim_x to c f$ exists, prove that $lim_x to c sqrtf = sqrt lim_x to c f$.
My Attempt:
Let us put $$ L colon= lim_x to c f(x). $$
Let us take any real number $varepsilon > 0$.
First suppose that $L = 0$. Then there exists a real number $delta > 0$ such that
$$ lvert f(x) - 0 rvert < varepsilon^2 $$
for all $x in A$ which satisfy
$$ 0 < lvert x - c rvert < delta. $$
But as $f(x) geq 0$ for all $x in A$, so for all $x in A$ for which
$$ 0 < lvert x-c rvert < delta, $$
we have
$$ f(x) = lvert f(x) rvert = lvert f(x) - 0 rvert < varepsilon^2, $$
and so
$$leftlvert left(sqrtfright)(x) - 0 rightrvert = sqrt f(x) < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = 0 $$
also.
Am I right?
Now suppose that $L neq 0$. If $L$ were negative, then we would have $-L > 0$ and so we could find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < -L $$
for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta. $$
Thus, for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta, $$
we would obtain
$$ L - (-L) < f(x) < L + (- L), $$
or
$$ 2L < f(x) < 0. $$
But $f(x) geq 0$ for all $x in A$. Thus if $L$ were negative, then we would arrive at a contradiction. So $L$ must be (strictly) positive.
Am I right?
Now as $L > 0$ and as $f(x) geq 0$ for all $x in A$, so for every point $x in A$, we find that
$$
beginalign
leftlvert left(sqrtfright)(x) - sqrtL rightrvert &= leftlvert sqrtf(x) - sqrtL rightrvert \
&= leftlvert frac left( sqrtf(x) - sqrtL right) left( sqrtf(x) + sqrtL right) sqrtf(x) + sqrtL rightrvert \
&= leftlvert frac f(x) - L sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert leftlvert sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert sqrtf(x) + sqrtL \
&leq frac leftlvert f(x) - L rightrvert sqrtL \
&= frac 1 sqrtL leftlvert f(x) - L rightrvert. tag1
endalign
$$
Now as $$ lim_x to c f(x) = L, $$
so we can find a real number $delta_0 > 0$ such that
$$ lvert f(x) - L rvert < varepsilon sqrtL $$
for all points $x in A$ for which $$ 0 < lvert x-c rvert < delta_0. $$
Thus from (1) we can conclude that, for all points $x in A$ for which $$ lvert x-crvert < delta_0, $$
we end up with
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert leq frac 1 sqrtL leftlvert f(x) - L rightrvert < frac 1 sqrtL varepsilon sqrtL = varepsilon, $$
and hence we obtain
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = sqrtL. $$
Am I right?
Is this proof good enough? What is the presentation like?
calculus real-analysis analysis limits proof-verification
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
add a comment |Â
up vote
1
down vote
favorite
Here is Prob. 15, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $A subseteq mathbbR$, let $f colon A to mathbbR$, and let $c in mathbbR$ be a cluster point of $A$. In addition, suppose that $f(x) geq 0$ for all $x in A$, and let $sqrtf$ be the function defined for $x in A$ by $left(sqrtfright)(x) = sqrtf(x)$. If $lim_x to c f$ exists, prove that $lim_x to c sqrtf = sqrt lim_x to c f$.
My Attempt:
Let us put $$ L colon= lim_x to c f(x). $$
Let us take any real number $varepsilon > 0$.
First suppose that $L = 0$. Then there exists a real number $delta > 0$ such that
$$ lvert f(x) - 0 rvert < varepsilon^2 $$
for all $x in A$ which satisfy
$$ 0 < lvert x - c rvert < delta. $$
But as $f(x) geq 0$ for all $x in A$, so for all $x in A$ for which
$$ 0 < lvert x-c rvert < delta, $$
we have
$$ f(x) = lvert f(x) rvert = lvert f(x) - 0 rvert < varepsilon^2, $$
and so
$$leftlvert left(sqrtfright)(x) - 0 rightrvert = sqrt f(x) < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = 0 $$
also.
Am I right?
Now suppose that $L neq 0$. If $L$ were negative, then we would have $-L > 0$ and so we could find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < -L $$
for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta. $$
Thus, for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta, $$
we would obtain
$$ L - (-L) < f(x) < L + (- L), $$
or
$$ 2L < f(x) < 0. $$
But $f(x) geq 0$ for all $x in A$. Thus if $L$ were negative, then we would arrive at a contradiction. So $L$ must be (strictly) positive.
Am I right?
Now as $L > 0$ and as $f(x) geq 0$ for all $x in A$, so for every point $x in A$, we find that
$$
beginalign
leftlvert left(sqrtfright)(x) - sqrtL rightrvert &= leftlvert sqrtf(x) - sqrtL rightrvert \
&= leftlvert frac left( sqrtf(x) - sqrtL right) left( sqrtf(x) + sqrtL right) sqrtf(x) + sqrtL rightrvert \
&= leftlvert frac f(x) - L sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert leftlvert sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert sqrtf(x) + sqrtL \
&leq frac leftlvert f(x) - L rightrvert sqrtL \
&= frac 1 sqrtL leftlvert f(x) - L rightrvert. tag1
endalign
$$
Now as $$ lim_x to c f(x) = L, $$
so we can find a real number $delta_0 > 0$ such that
$$ lvert f(x) - L rvert < varepsilon sqrtL $$
for all points $x in A$ for which $$ 0 < lvert x-c rvert < delta_0. $$
Thus from (1) we can conclude that, for all points $x in A$ for which $$ lvert x-crvert < delta_0, $$
we end up with
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert leq frac 1 sqrtL leftlvert f(x) - L rightrvert < frac 1 sqrtL varepsilon sqrtL = varepsilon, $$
and hence we obtain
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = sqrtL. $$
Am I right?
Is this proof good enough? What is the presentation like?
calculus real-analysis analysis limits proof-verification
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
I didn't read everything in detail but looks good. Another way is to prove first that if a function is always $ge 0$, then any limit is also $ge 0$.
– Ovi
Aug 3 at 8:14
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is Prob. 15, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $A subseteq mathbbR$, let $f colon A to mathbbR$, and let $c in mathbbR$ be a cluster point of $A$. In addition, suppose that $f(x) geq 0$ for all $x in A$, and let $sqrtf$ be the function defined for $x in A$ by $left(sqrtfright)(x) = sqrtf(x)$. If $lim_x to c f$ exists, prove that $lim_x to c sqrtf = sqrt lim_x to c f$.
My Attempt:
Let us put $$ L colon= lim_x to c f(x). $$
Let us take any real number $varepsilon > 0$.
First suppose that $L = 0$. Then there exists a real number $delta > 0$ such that
$$ lvert f(x) - 0 rvert < varepsilon^2 $$
for all $x in A$ which satisfy
$$ 0 < lvert x - c rvert < delta. $$
But as $f(x) geq 0$ for all $x in A$, so for all $x in A$ for which
$$ 0 < lvert x-c rvert < delta, $$
we have
$$ f(x) = lvert f(x) rvert = lvert f(x) - 0 rvert < varepsilon^2, $$
and so
$$leftlvert left(sqrtfright)(x) - 0 rightrvert = sqrt f(x) < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = 0 $$
also.
Am I right?
Now suppose that $L neq 0$. If $L$ were negative, then we would have $-L > 0$ and so we could find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < -L $$
for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta. $$
Thus, for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta, $$
we would obtain
$$ L - (-L) < f(x) < L + (- L), $$
or
$$ 2L < f(x) < 0. $$
But $f(x) geq 0$ for all $x in A$. Thus if $L$ were negative, then we would arrive at a contradiction. So $L$ must be (strictly) positive.
Am I right?
Now as $L > 0$ and as $f(x) geq 0$ for all $x in A$, so for every point $x in A$, we find that
$$
beginalign
leftlvert left(sqrtfright)(x) - sqrtL rightrvert &= leftlvert sqrtf(x) - sqrtL rightrvert \
&= leftlvert frac left( sqrtf(x) - sqrtL right) left( sqrtf(x) + sqrtL right) sqrtf(x) + sqrtL rightrvert \
&= leftlvert frac f(x) - L sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert leftlvert sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert sqrtf(x) + sqrtL \
&leq frac leftlvert f(x) - L rightrvert sqrtL \
&= frac 1 sqrtL leftlvert f(x) - L rightrvert. tag1
endalign
$$
Now as $$ lim_x to c f(x) = L, $$
so we can find a real number $delta_0 > 0$ such that
$$ lvert f(x) - L rvert < varepsilon sqrtL $$
for all points $x in A$ for which $$ 0 < lvert x-c rvert < delta_0. $$
Thus from (1) we can conclude that, for all points $x in A$ for which $$ lvert x-crvert < delta_0, $$
we end up with
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert leq frac 1 sqrtL leftlvert f(x) - L rightrvert < frac 1 sqrtL varepsilon sqrtL = varepsilon, $$
and hence we obtain
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = sqrtL. $$
Am I right?
Is this proof good enough? What is the presentation like?
calculus real-analysis analysis limits proof-verification
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
Here is Prob. 15, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $A subseteq mathbbR$, let $f colon A to mathbbR$, and let $c in mathbbR$ be a cluster point of $A$. In addition, suppose that $f(x) geq 0$ for all $x in A$, and let $sqrtf$ be the function defined for $x in A$ by $left(sqrtfright)(x) = sqrtf(x)$. If $lim_x to c f$ exists, prove that $lim_x to c sqrtf = sqrt lim_x to c f$.
My Attempt:
Let us put $$ L colon= lim_x to c f(x). $$
Let us take any real number $varepsilon > 0$.
First suppose that $L = 0$. Then there exists a real number $delta > 0$ such that
$$ lvert f(x) - 0 rvert < varepsilon^2 $$
for all $x in A$ which satisfy
$$ 0 < lvert x - c rvert < delta. $$
But as $f(x) geq 0$ for all $x in A$, so for all $x in A$ for which
$$ 0 < lvert x-c rvert < delta, $$
we have
$$ f(x) = lvert f(x) rvert = lvert f(x) - 0 rvert < varepsilon^2, $$
and so
$$leftlvert left(sqrtfright)(x) - 0 rightrvert = sqrt f(x) < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = 0 $$
also.
Am I right?
Now suppose that $L neq 0$. If $L$ were negative, then we would have $-L > 0$ and so we could find a real number $delta > 0$ such that
$$ lvert f(x) - L rvert < -L $$
for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta. $$
Thus, for all $x in A$ which satisfy
$$ 0 < lvert x-crvert < delta, $$
we would obtain
$$ L - (-L) < f(x) < L + (- L), $$
or
$$ 2L < f(x) < 0. $$
But $f(x) geq 0$ for all $x in A$. Thus if $L$ were negative, then we would arrive at a contradiction. So $L$ must be (strictly) positive.
Am I right?
Now as $L > 0$ and as $f(x) geq 0$ for all $x in A$, so for every point $x in A$, we find that
$$
beginalign
leftlvert left(sqrtfright)(x) - sqrtL rightrvert &= leftlvert sqrtf(x) - sqrtL rightrvert \
&= leftlvert frac left( sqrtf(x) - sqrtL right) left( sqrtf(x) + sqrtL right) sqrtf(x) + sqrtL rightrvert \
&= leftlvert frac f(x) - L sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert leftlvert sqrtf(x) + sqrtL rightrvert \
&= frac leftlvert f(x) - L rightrvert sqrtf(x) + sqrtL \
&leq frac leftlvert f(x) - L rightrvert sqrtL \
&= frac 1 sqrtL leftlvert f(x) - L rightrvert. tag1
endalign
$$
Now as $$ lim_x to c f(x) = L, $$
so we can find a real number $delta_0 > 0$ such that
$$ lvert f(x) - L rvert < varepsilon sqrtL $$
for all points $x in A$ for which $$ 0 < lvert x-c rvert < delta_0. $$
Thus from (1) we can conclude that, for all points $x in A$ for which $$ lvert x-crvert < delta_0, $$
we end up with
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert leq frac 1 sqrtL leftlvert f(x) - L rightrvert < frac 1 sqrtL varepsilon sqrtL = varepsilon, $$
and hence we obtain
$$ leftlvert left(sqrtfright)(x) - sqrtL rightrvert < varepsilon. $$
Thus it follows that
$$ lim_x to c left(sqrtfright)(x) = sqrtL. $$
Am I right?
Is this proof good enough? What is the presentation like?
calculus real-analysis analysis limits proof-verification
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
asked Aug 3 at 8:02
Saaqib Mahmood
7,09542169
7,09542169
I didn't read everything in detail but looks good. Another way is to prove first that if a function is always $ge 0$, then any limit is also $ge 0$.
– Ovi
Aug 3 at 8:14
add a comment |Â
I didn't read everything in detail but looks good. Another way is to prove first that if a function is always $ge 0$, then any limit is also $ge 0$.
– Ovi
Aug 3 at 8:14
I didn't read everything in detail but looks good. Another way is to prove first that if a function is always $ge 0$, then any limit is also $ge 0$.
– Ovi
Aug 3 at 8:14
I didn't read everything in detail but looks good. Another way is to prove first that if a function is always $ge 0$, then any limit is also $ge 0$.
– Ovi
Aug 3 at 8:14
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870853%2fprob-15-sec-4-2-in-bartle-sherberts-intro-to-real-analysis-lim-x-to-c%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I didn't read everything in detail but looks good. Another way is to prove first that if a function is always $ge 0$, then any limit is also $ge 0$.
– Ovi
Aug 3 at 8:14