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Prob. 5 (e), Sec. 4.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to find $lim_xto 0- fracsqrtx+1x$?
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up vote
1
down vote
favorite
I know that
$$ lim_x to 0+ frac sqrt x+1 x = +infty. $$
Now how to find
$$ lim_x to 0- frac sqrt x+1 x = +infty? $$
This is Prob. 5 (e), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.
My Attempt:
The function $f$ given by
$$ f(x) colon= frac sqrt x+1 x $$
is defined only for real numbers $x geq -1$. Moreover, $f(x) < 0$ for $x < 0$.
If $-1 < x < 0$, then
$$ 0 < sqrt x+1 < 1, $$
and so
$$ frac1x < frac sqrt x+1 x < 0. tag1 $$
Also if $-1 < x < 0$, then
$$ -frac1x > 1, $$
and as $sqrtx+1 > 0$, so
$$ - frac sqrt x+1 x > sqrtx+1, $$
and hence
$$ frac sqrt x+1 x < - sqrtx+1. $$
What to do here?
I've really no clue of what to do. I would like to be able to majorise this function by one tending to $-infty$ as $x to 0-$, or I would like to bound this function between two functions that both have the same limit as $x to 0-$.
Or, does the limit exist at all in the extended real number system $mathbbR cup left pm infty right$?
calculus real-analysis limits analysis
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
add a comment |Â
up vote
1
down vote
favorite
I know that
$$ lim_x to 0+ frac sqrt x+1 x = +infty. $$
Now how to find
$$ lim_x to 0- frac sqrt x+1 x = +infty? $$
This is Prob. 5 (e), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.
My Attempt:
The function $f$ given by
$$ f(x) colon= frac sqrt x+1 x $$
is defined only for real numbers $x geq -1$. Moreover, $f(x) < 0$ for $x < 0$.
If $-1 < x < 0$, then
$$ 0 < sqrt x+1 < 1, $$
and so
$$ frac1x < frac sqrt x+1 x < 0. tag1 $$
Also if $-1 < x < 0$, then
$$ -frac1x > 1, $$
and as $sqrtx+1 > 0$, so
$$ - frac sqrt x+1 x > sqrtx+1, $$
and hence
$$ frac sqrt x+1 x < - sqrtx+1. $$
What to do here?
I've really no clue of what to do. I would like to be able to majorise this function by one tending to $-infty$ as $x to 0-$, or I would like to bound this function between two functions that both have the same limit as $x to 0-$.
Or, does the limit exist at all in the extended real number system $mathbbR cup left pm infty right$?
calculus real-analysis limits analysis
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
I think you are wrong, it should be $-infty$.
– mathreadler
Aug 6 at 15:40
Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-infty$. To show that rigoursly we can use for example squeeze theorem.
– gimusi
Aug 6 at 15:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that
$$ lim_x to 0+ frac sqrt x+1 x = +infty. $$
Now how to find
$$ lim_x to 0- frac sqrt x+1 x = +infty? $$
This is Prob. 5 (e), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.
My Attempt:
The function $f$ given by
$$ f(x) colon= frac sqrt x+1 x $$
is defined only for real numbers $x geq -1$. Moreover, $f(x) < 0$ for $x < 0$.
If $-1 < x < 0$, then
$$ 0 < sqrt x+1 < 1, $$
and so
$$ frac1x < frac sqrt x+1 x < 0. tag1 $$
Also if $-1 < x < 0$, then
$$ -frac1x > 1, $$
and as $sqrtx+1 > 0$, so
$$ - frac sqrt x+1 x > sqrtx+1, $$
and hence
$$ frac sqrt x+1 x < - sqrtx+1. $$
What to do here?
I've really no clue of what to do. I would like to be able to majorise this function by one tending to $-infty$ as $x to 0-$, or I would like to bound this function between two functions that both have the same limit as $x to 0-$.
Or, does the limit exist at all in the extended real number system $mathbbR cup left pm infty right$?
calculus real-analysis limits analysis
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
I know that
$$ lim_x to 0+ frac sqrt x+1 x = +infty. $$
Now how to find
$$ lim_x to 0- frac sqrt x+1 x = +infty? $$
This is Prob. 5 (e), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.
My Attempt:
The function $f$ given by
$$ f(x) colon= frac sqrt x+1 x $$
is defined only for real numbers $x geq -1$. Moreover, $f(x) < 0$ for $x < 0$.
If $-1 < x < 0$, then
$$ 0 < sqrt x+1 < 1, $$
and so
$$ frac1x < frac sqrt x+1 x < 0. tag1 $$
Also if $-1 < x < 0$, then
$$ -frac1x > 1, $$
and as $sqrtx+1 > 0$, so
$$ - frac sqrt x+1 x > sqrtx+1, $$
and hence
$$ frac sqrt x+1 x < - sqrtx+1. $$
What to do here?
I've really no clue of what to do. I would like to be able to majorise this function by one tending to $-infty$ as $x to 0-$, or I would like to bound this function between two functions that both have the same limit as $x to 0-$.
Or, does the limit exist at all in the extended real number system $mathbbR cup left pm infty right$?
calculus real-analysis limits analysis
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
edited Aug 6 at 15:44
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
asked Aug 6 at 15:33
Saaqib Mahmood
7,17542169
7,17542169
I think you are wrong, it should be $-infty$.
– mathreadler
Aug 6 at 15:40
Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-infty$. To show that rigoursly we can use for example squeeze theorem.
– gimusi
Aug 6 at 15:45
add a comment |Â
I think you are wrong, it should be $-infty$.
– mathreadler
Aug 6 at 15:40
Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-infty$. To show that rigoursly we can use for example squeeze theorem.
– gimusi
Aug 6 at 15:45
I think you are wrong, it should be $-infty$.
– mathreadler
Aug 6 at 15:40
I think you are wrong, it should be $-infty$.
– mathreadler
Aug 6 at 15:40
Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-infty$. To show that rigoursly we can use for example squeeze theorem.
– gimusi
Aug 6 at 15:45
Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-infty$. To show that rigoursly we can use for example squeeze theorem.
– gimusi
Aug 6 at 15:45
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$$ fracsqrtx + 1x
=fracx + 1xsqrtx + 1
=frac1sqrtx + 1 + frac1xsqrtx + 1
to -infty quad textas quad x to 0^-
$$
answered Aug 6 at 15:54
steven gregory
16.5k22055
Very nice approach!
– gimusi
Aug 6 at 16:06
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
add a comment |Â
up vote
3
down vote
We have that as $xto 0^+$
$$frac sqrt x+1 xge frac 1 x to infty$$
and as $xto 0^-$ let $y=-x to 0^+$ eventually as $sqrt 1-y ge frac12$
$$frac sqrt x+1 x=-frac sqrt 1-y yle -frac 1 2y to -infty$$
therefore the limit doesn't exist but the one side limits exist in $mathbbR cup left pm infty right$.
answered Aug 6 at 15:36
gimusi
65.4k73684
add a comment |Â
up vote
2
down vote
The stated result isn't right. $1/x to -infty$ as $xto 0^-$. The numerator is bounded and positive near $x=0$, so the answer remains the same. A full proof: Since $sqrt1+x$ is increasing in $x$,
$$ fracsqrt1+xx ≤ fracsqrt1+0x = frac1x to -infty quad (xto 0^-) $$
answered Aug 6 at 15:41
Calvin Khor
8,18411133
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$ fracsqrtx + 1x
=fracx + 1xsqrtx + 1
=frac1sqrtx + 1 + frac1xsqrtx + 1
to -infty quad textas quad x to 0^-
$$
answered Aug 6 at 15:54
steven gregory
16.5k22055
Very nice approach!
– gimusi
Aug 6 at 16:06
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
add a comment |Â
up vote
1
down vote
accepted
$$ fracsqrtx + 1x
=fracx + 1xsqrtx + 1
=frac1sqrtx + 1 + frac1xsqrtx + 1
to -infty quad textas quad x to 0^-
$$
answered Aug 6 at 15:54
steven gregory
16.5k22055
Very nice approach!
– gimusi
Aug 6 at 16:06
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$ fracsqrtx + 1x
=fracx + 1xsqrtx + 1
=frac1sqrtx + 1 + frac1xsqrtx + 1
to -infty quad textas quad x to 0^-
$$
answered Aug 6 at 15:54
steven gregory
16.5k22055
$$ fracsqrtx + 1x
=fracx + 1xsqrtx + 1
=frac1sqrtx + 1 + frac1xsqrtx + 1
to -infty quad textas quad x to 0^-
$$
answered Aug 6 at 15:54
steven gregory
16.5k22055
answered Aug 6 at 15:54
steven gregory
16.5k22055
answered Aug 6 at 15:54
steven gregory
16.5k22055
answered Aug 6 at 15:54
steven gregory
16.5k22055
16.5k22055
Very nice approach!
– gimusi
Aug 6 at 16:06
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
add a comment |Â
Very nice approach!
– gimusi
Aug 6 at 16:06
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
Very nice approach!
– gimusi
Aug 6 at 16:06
Very nice approach!
– gimusi
Aug 6 at 16:06
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
@stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own!
– Saaqib Mahmood
Aug 6 at 17:33
add a comment |Â
up vote
3
down vote
We have that as $xto 0^+$
$$frac sqrt x+1 xge frac 1 x to infty$$
and as $xto 0^-$ let $y=-x to 0^+$ eventually as $sqrt 1-y ge frac12$
$$frac sqrt x+1 x=-frac sqrt 1-y yle -frac 1 2y to -infty$$
therefore the limit doesn't exist but the one side limits exist in $mathbbR cup left pm infty right$.
answered Aug 6 at 15:36
gimusi
65.4k73684
add a comment |Â
up vote
3
down vote
We have that as $xto 0^+$
$$frac sqrt x+1 xge frac 1 x to infty$$
and as $xto 0^-$ let $y=-x to 0^+$ eventually as $sqrt 1-y ge frac12$
$$frac sqrt x+1 x=-frac sqrt 1-y yle -frac 1 2y to -infty$$
therefore the limit doesn't exist but the one side limits exist in $mathbbR cup left pm infty right$.
answered Aug 6 at 15:36
gimusi
65.4k73684
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have that as $xto 0^+$
$$frac sqrt x+1 xge frac 1 x to infty$$
and as $xto 0^-$ let $y=-x to 0^+$ eventually as $sqrt 1-y ge frac12$
$$frac sqrt x+1 x=-frac sqrt 1-y yle -frac 1 2y to -infty$$
therefore the limit doesn't exist but the one side limits exist in $mathbbR cup left pm infty right$.
answered Aug 6 at 15:36
gimusi
65.4k73684
We have that as $xto 0^+$
$$frac sqrt x+1 xge frac 1 x to infty$$
and as $xto 0^-$ let $y=-x to 0^+$ eventually as $sqrt 1-y ge frac12$
$$frac sqrt x+1 x=-frac sqrt 1-y yle -frac 1 2y to -infty$$
therefore the limit doesn't exist but the one side limits exist in $mathbbR cup left pm infty right$.
answered Aug 6 at 15:36
gimusi
65.4k73684
answered Aug 6 at 15:36
gimusi
65.4k73684
answered Aug 6 at 15:36
gimusi
65.4k73684
answered Aug 6 at 15:36
gimusi
65.4k73684
65.4k73684
add a comment |Â
add a comment |Â
up vote
2
down vote
The stated result isn't right. $1/x to -infty$ as $xto 0^-$. The numerator is bounded and positive near $x=0$, so the answer remains the same. A full proof: Since $sqrt1+x$ is increasing in $x$,
$$ fracsqrt1+xx ≤ fracsqrt1+0x = frac1x to -infty quad (xto 0^-) $$
answered Aug 6 at 15:41
Calvin Khor
8,18411133
add a comment |Â
up vote
2
down vote
The stated result isn't right. $1/x to -infty$ as $xto 0^-$. The numerator is bounded and positive near $x=0$, so the answer remains the same. A full proof: Since $sqrt1+x$ is increasing in $x$,
$$ fracsqrt1+xx ≤ fracsqrt1+0x = frac1x to -infty quad (xto 0^-) $$
answered Aug 6 at 15:41
Calvin Khor
8,18411133
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The stated result isn't right. $1/x to -infty$ as $xto 0^-$. The numerator is bounded and positive near $x=0$, so the answer remains the same. A full proof: Since $sqrt1+x$ is increasing in $x$,
$$ fracsqrt1+xx ≤ fracsqrt1+0x = frac1x to -infty quad (xto 0^-) $$
answered Aug 6 at 15:41
Calvin Khor
8,18411133
The stated result isn't right. $1/x to -infty$ as $xto 0^-$. The numerator is bounded and positive near $x=0$, so the answer remains the same. A full proof: Since $sqrt1+x$ is increasing in $x$,
$$ fracsqrt1+xx ≤ fracsqrt1+0x = frac1x to -infty quad (xto 0^-) $$
answered Aug 6 at 15:41
Calvin Khor
8,18411133
answered Aug 6 at 15:41
Calvin Khor
8,18411133
answered Aug 6 at 15:41
Calvin Khor
8,18411133
answered Aug 6 at 15:41
Calvin Khor
8,18411133
8,18411133
add a comment |Â
add a comment |Â
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I think you are wrong, it should be $-infty$.
– mathreadler
Aug 6 at 15:40
Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-infty$. To show that rigoursly we can use for example squeeze theorem.
– gimusi
Aug 6 at 15:45