Prob. 5 (b), Sec. 4.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Rigorous Proof of a Limit Statement

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Let $f colon mathbbR setminus 1 to mathbbR$ be the function defined by
$$ f(x) colon= fracxx-1 mbox for all x in mathbbR setminus 1 . $$
Then how to prove rigorously the following limit statement?
$$ lim_x to 1 - f(x) = -infty. $$

This is Prob. 5 (b), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.

Here is my attempt at a rigorous proof of the above statement.

Let us first restrict our $x$ such that $$ 0 < x < 1. tag0 $$
Then $x-1 < 0$ and so $$ fracxx-1 < 0. tag1 $$

Now let us take any real number $alpha$.

CASE 1: If $alpha geq 0$, then we find from (1) that $f(x) < alpha$
for all $x in mathbbR$ such that $0 < x < 1$.

So if we choose a real number $delta$ such that $0 < delta < 1$, then $-1 < -delta$ and hence $0 < 1-delta$, and thus for all $x in mathbbRsetminus 1 $ such that
$ 1 - delta < x < 1$, we have $0 < x < 1$ which implies that $f(x) < alpha$.

CASE 2: Now let us take our $alpha < 0$.

Then for all $x in mathbbR$ for which (0) above holds, we see that $f(x) < alpha$ holds if and only if
$$ fracxx-1 < alpha, $$
and this holds if and only if
$$ x > (x-1) alpha, $$
[ because $x-1 < 0$, by virtue of (0) ]
or
$$ x > x alpha - alpha tag2 $$

Now (2) holds if and only if
$$ alpha > x alpha - x = x(alpha - 1), $$
and this holds if and only if
$$ fracalphaalpha - 1 < x, tag3 $$
because $alpha - 1 < 0$. [ Note that here we have assumed that $alpha < 0$. ]

Now the left-hand side of (3) can be rewritten as
$$ 1 - frac11-alpha < x. tag3 $$

Now let us choose any real number $delta$ in such a manner that
$$ 0 < delta leq frac11-alpha. $$
Then
$$ - frac11-alpha leq - delta, $$
and hence
$$ 1 - frac11- alpha leq 1 - delta. tag4 $$

So for all $x in mathbbR$, if
$$ 1 - delta < x < 1, $$
then from (4) we can conclude that for all those $x$ we also have
$$ 1 - frac11- alpha < x < 1, $$
and so from (0) and (3) we obtain
$$ f(x) < alpha. $$

Thus in either case we have shown that, for every real number $alpha$, we can find a real number $delta > 0$ so that $f(x) < alpha$ for all $x in mathbbR$ which satisfy $$ 1-delta < x < 1. $$

Hence $$ lim_x to 1- fracxx-1 = - infty, $$
as required.

Is this proof correct and clear enough? If not, then where are the deficiencies?

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

add a commentÂ |Â

up vote
1
down vote

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This is Prob. 5 (b), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.

Here is my attempt at a rigorous proof of the above statement.

Let us first restrict our $x$ such that $$ 0 < x < 1. tag0 $$
Then $x-1 < 0$ and so $$ fracxx-1 < 0. tag1 $$

Now let us take any real number $alpha$.

CASE 1: If $alpha geq 0$, then we find from (1) that $f(x) < alpha$
for all $x in mathbbR$ such that $0 < x < 1$.

So if we choose a real number $delta$ such that $0 < delta < 1$, then $-1 < -delta$ and hence $0 < 1-delta$, and thus for all $x in mathbbRsetminus 1 $ such that
$ 1 - delta < x < 1$, we have $0 < x < 1$ which implies that $f(x) < alpha$.

CASE 2: Now let us take our $alpha < 0$.

Then for all $x in mathbbR$ for which (0) above holds, we see that $f(x) < alpha$ holds if and only if
$$ fracxx-1 < alpha, $$
and this holds if and only if
$$ x > (x-1) alpha, $$
[ because $x-1 < 0$, by virtue of (0) ]
or
$$ x > x alpha - alpha tag2 $$

Now (2) holds if and only if
$$ alpha > x alpha - x = x(alpha - 1), $$
and this holds if and only if
$$ fracalphaalpha - 1 < x, tag3 $$
because $alpha - 1 < 0$. [ Note that here we have assumed that $alpha < 0$. ]

Now the left-hand side of (3) can be rewritten as
$$ 1 - frac11-alpha < x. tag3 $$

Now let us choose any real number $delta$ in such a manner that
$$ 0 < delta leq frac11-alpha. $$
Then
$$ - frac11-alpha leq - delta, $$
and hence
$$ 1 - frac11- alpha leq 1 - delta. tag4 $$

So for all $x in mathbbR$, if
$$ 1 - delta < x < 1, $$
then from (4) we can conclude that for all those $x$ we also have
$$ 1 - frac11- alpha < x < 1, $$
and so from (0) and (3) we obtain
$$ f(x) < alpha. $$

Thus in either case we have shown that, for every real number $alpha$, we can find a real number $delta > 0$ so that $f(x) < alpha$ for all $x in mathbbR$ which satisfy $$ 1-delta < x < 1. $$

Hence $$ lim_x to 1- fracxx-1 = - infty, $$
as required.

Is this proof correct and clear enough? If not, then where are the deficiencies?

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

add a commentÂ |Â

up vote
1
down vote

favorite

This is Prob. 5 (b), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.

Here is my attempt at a rigorous proof of the above statement.

Let us first restrict our $x$ such that $$ 0 < x < 1. tag0 $$
Then $x-1 < 0$ and so $$ fracxx-1 < 0. tag1 $$

Now let us take any real number $alpha$.

CASE 1: If $alpha geq 0$, then we find from (1) that $f(x) < alpha$
for all $x in mathbbR$ such that $0 < x < 1$.

So if we choose a real number $delta$ such that $0 < delta < 1$, then $-1 < -delta$ and hence $0 < 1-delta$, and thus for all $x in mathbbRsetminus 1 $ such that
$ 1 - delta < x < 1$, we have $0 < x < 1$ which implies that $f(x) < alpha$.

CASE 2: Now let us take our $alpha < 0$.

Then for all $x in mathbbR$ for which (0) above holds, we see that $f(x) < alpha$ holds if and only if
$$ fracxx-1 < alpha, $$
and this holds if and only if
$$ x > (x-1) alpha, $$
[ because $x-1 < 0$, by virtue of (0) ]
or
$$ x > x alpha - alpha tag2 $$

Now (2) holds if and only if
$$ alpha > x alpha - x = x(alpha - 1), $$
and this holds if and only if
$$ fracalphaalpha - 1 < x, tag3 $$
because $alpha - 1 < 0$. [ Note that here we have assumed that $alpha < 0$. ]

Now the left-hand side of (3) can be rewritten as
$$ 1 - frac11-alpha < x. tag3 $$

Now let us choose any real number $delta$ in such a manner that
$$ 0 < delta leq frac11-alpha. $$
Then
$$ - frac11-alpha leq - delta, $$
and hence
$$ 1 - frac11- alpha leq 1 - delta. tag4 $$

So for all $x in mathbbR$, if
$$ 1 - delta < x < 1, $$
then from (4) we can conclude that for all those $x$ we also have
$$ 1 - frac11- alpha < x < 1, $$
and so from (0) and (3) we obtain
$$ f(x) < alpha. $$

Thus in either case we have shown that, for every real number $alpha$, we can find a real number $delta > 0$ so that $f(x) < alpha$ for all $x in mathbbR$ which satisfy $$ 1-delta < x < 1. $$

Hence $$ lim_x to 1- fracxx-1 = - infty, $$
as required.

Is this proof correct and clear enough? If not, then where are the deficiencies?

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

This is Prob. 5 (b), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.

Here is my attempt at a rigorous proof of the above statement.

Let us first restrict our $x$ such that $$ 0 < x < 1. tag0 $$
Then $x-1 < 0$ and so $$ fracxx-1 < 0. tag1 $$

Now let us take any real number $alpha$.

CASE 1: If $alpha geq 0$, then we find from (1) that $f(x) < alpha$
for all $x in mathbbR$ such that $0 < x < 1$.

So if we choose a real number $delta$ such that $0 < delta < 1$, then $-1 < -delta$ and hence $0 < 1-delta$, and thus for all $x in mathbbRsetminus 1 $ such that
$ 1 - delta < x < 1$, we have $0 < x < 1$ which implies that $f(x) < alpha$.

CASE 2: Now let us take our $alpha < 0$.

Then for all $x in mathbbR$ for which (0) above holds, we see that $f(x) < alpha$ holds if and only if
$$ fracxx-1 < alpha, $$
and this holds if and only if
$$ x > (x-1) alpha, $$
[ because $x-1 < 0$, by virtue of (0) ]
or
$$ x > x alpha - alpha tag2 $$

Now (2) holds if and only if
$$ alpha > x alpha - x = x(alpha - 1), $$
and this holds if and only if
$$ fracalphaalpha - 1 < x, tag3 $$
because $alpha - 1 < 0$. [ Note that here we have assumed that $alpha < 0$. ]

Now the left-hand side of (3) can be rewritten as
$$ 1 - frac11-alpha < x. tag3 $$

Now let us choose any real number $delta$ in such a manner that
$$ 0 < delta leq frac11-alpha. $$
Then
$$ - frac11-alpha leq - delta, $$
and hence
$$ 1 - frac11- alpha leq 1 - delta. tag4 $$

So for all $x in mathbbR$, if
$$ 1 - delta < x < 1, $$
then from (4) we can conclude that for all those $x$ we also have
$$ 1 - frac11- alpha < x < 1, $$
and so from (0) and (3) we obtain
$$ f(x) < alpha. $$

Thus in either case we have shown that, for every real number $alpha$, we can find a real number $delta > 0$ so that $f(x) < alpha$ for all $x in mathbbR$ which satisfy $$ 1-delta < x < 1. $$

Hence $$ lim_x to 1- fracxx-1 = - infty, $$
as required.

Is this proof correct and clear enough? If not, then where are the deficiencies?

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

asked Aug 6 at 12:58

Saaqib Mahmood

7,17542169

add a commentÂ |Â

1 Answer
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up vote
0
down vote

It seems correct, but it's too long and verbose.

You have to prove that, for every $alpha>0$, there exists $delta>0$ such that, for $1-delta<x<1$,
$$
fracxx-1<-alpha
$$
This can be rewritten as
$$
fracx(1+alpha)-alphax-1<0
$$
which is satisfied for
$$
fracalpha1+alpha<x<1
$$
Note that
$$
fracalpha1+alpha=1-frac11+alpha
$$
and take $delta=1/(1+alpha)$.

You don't even discuss the case $alphale0$; if you're pedantic, just take $delta=1/2$ for every $alphale 0$ (the value of $delta$ corresponding to $alpha=1$.

edited Aug 6 at 13:56

answered Aug 6 at 13:38

egreg

164k1180187

thank you for your solution. As you've rightly pointed out, my proof is indeed rather long and verbose. Actually, I've tried to show in detail exactly how I've come up with the $delta$ which I have. But you've proved the limit statement for a slightly different function.
â€“Â Saaqib Mahmood
Aug 6 at 13:46

@SaaqibMahmood Oh, well; here's an even easier proof.
â€“Â egreg
Aug 6 at 13:57

add a commentÂ |Â

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1 Answer
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up vote
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down vote

It seems correct, but it's too long and verbose.

You don't even discuss the case $alphale0$; if you're pedantic, just take $delta=1/2$ for every $alphale 0$ (the value of $delta$ corresponding to $alpha=1$.

edited Aug 6 at 13:56

answered Aug 6 at 13:38

egreg

164k1180187

thank you for your solution. As you've rightly pointed out, my proof is indeed rather long and verbose. Actually, I've tried to show in detail exactly how I've come up with the $delta$ which I have. But you've proved the limit statement for a slightly different function.
â€“Â Saaqib Mahmood
Aug 6 at 13:46

@SaaqibMahmood Oh, well; here's an even easier proof.
â€“Â egreg
Aug 6 at 13:57

add a commentÂ |Â

up vote
0
down vote

It seems correct, but it's too long and verbose.

You don't even discuss the case $alphale0$; if you're pedantic, just take $delta=1/2$ for every $alphale 0$ (the value of $delta$ corresponding to $alpha=1$.

edited Aug 6 at 13:56

answered Aug 6 at 13:38

egreg

164k1180187

thank you for your solution. As you've rightly pointed out, my proof is indeed rather long and verbose. Actually, I've tried to show in detail exactly how I've come up with the $delta$ which I have. But you've proved the limit statement for a slightly different function.
â€“Â Saaqib Mahmood
Aug 6 at 13:46

@SaaqibMahmood Oh, well; here's an even easier proof.
â€“Â egreg
Aug 6 at 13:57

add a commentÂ |Â

up vote
0
down vote

It seems correct, but it's too long and verbose.

You don't even discuss the case $alphale0$; if you're pedantic, just take $delta=1/2$ for every $alphale 0$ (the value of $delta$ corresponding to $alpha=1$.

edited Aug 6 at 13:56

answered Aug 6 at 13:38

egreg

164k1180187

It seems correct, but it's too long and verbose.

You don't even discuss the case $alphale0$; if you're pedantic, just take $delta=1/2$ for every $alphale 0$ (the value of $delta$ corresponding to $alpha=1$.

edited Aug 6 at 13:56

answered Aug 6 at 13:38

egreg

164k1180187

edited Aug 6 at 13:56

answered Aug 6 at 13:38

egreg

164k1180187

answered Aug 6 at 13:38

egreg

164k1180187

answered Aug 6 at 13:38

egreg

164k1180187

thank you for your solution. As you've rightly pointed out, my proof is indeed rather long and verbose. Actually, I've tried to show in detail exactly how I've come up with the $delta$ which I have. But you've proved the limit statement for a slightly different function.
â€“Â Saaqib Mahmood
Aug 6 at 13:46

@SaaqibMahmood Oh, well; here's an even easier proof.
â€“Â egreg
Aug 6 at 13:57

add a commentÂ |Â

thank you for your solution. As you've rightly pointed out, my proof is indeed rather long and verbose. Actually, I've tried to show in detail exactly how I've come up with the $delta$ which I have. But you've proved the limit statement for a slightly different function.
â€“Â Saaqib Mahmood
Aug 6 at 13:46

@SaaqibMahmood Oh, well; here's an even easier proof.
â€“Â egreg
Aug 6 at 13:57

thank you for your solution. As you've rightly pointed out, my proof is indeed rather long and verbose. Actually, I've tried to show in detail exactly how I've come up with the $delta$ which I have. But you've proved the limit statement for a slightly different function.
â€“Â Saaqib Mahmood
Aug 6 at 13:46

@SaaqibMahmood Oh, well; here's an even easier proof.
â€“Â egreg
Aug 6 at 13:57

add a commentÂ |Â

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