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Find $lim_xrightarrow 1^pmbig(frac 2x+3x^2-1big)$
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I want to find
$$lim_xrightarrow 1^pmleft(frac 2x+3x^2-1right)$$
This is what my textbook does:
$$frac2x+3x^2-1=frac2x+3(x-1)(x+1)=frac 1 x-1cdotfrac2x+3x+1$$
Now, for $xrightarrow1$, we have
$$frac2x+3x+1rightarrow frac 5 2$$
For $xrightarrow1^pm$ we have
$$frac1x-1rightarrowpminfty$$
This means that
$$frac 2x+3x^2-1 rightarrow pminfty$$
This is indeed correct. What I instead did was this:
$$frac 2x+3x^2-1sim frac2xx^2 = frac 2x$$
Then I would "replace" (not sure if this is something I can do) $x$ with $1$ (it's the number which the limit approaches to):
$$frac 2 1 rightarrow2$$
The limit would be $2$. Replacing $x$ with the number that the limit approaches I think is something I learnt in high school. Any hints on why my reasoning is wrong?
limits
edited Aug 6 at 15:13
Daniel Buck
2,3341624
asked Aug 6 at 14:53
Cesare
42129
 |Â
show 8 more comments
up vote
0
down vote
favorite
I want to find
$$lim_xrightarrow 1^pmleft(frac 2x+3x^2-1right)$$
This is what my textbook does:
$$frac2x+3x^2-1=frac2x+3(x-1)(x+1)=frac 1 x-1cdotfrac2x+3x+1$$
Now, for $xrightarrow1$, we have
$$frac2x+3x+1rightarrow frac 5 2$$
For $xrightarrow1^pm$ we have
$$frac1x-1rightarrowpminfty$$
This means that
$$frac 2x+3x^2-1 rightarrow pminfty$$
This is indeed correct. What I instead did was this:
$$frac 2x+3x^2-1sim frac2xx^2 = frac 2x$$
Then I would "replace" (not sure if this is something I can do) $x$ with $1$ (it's the number which the limit approaches to):
$$frac 2 1 rightarrow2$$
The limit would be $2$. Replacing $x$ with the number that the limit approaches I think is something I learnt in high school. Any hints on why my reasoning is wrong?
limits
edited Aug 6 at 15:13
Daniel Buck
2,3341624
asked Aug 6 at 14:53
Cesare
42129
How does $x$ depend on $n$?
– Mark Viola
Aug 6 at 14:55
1
What is $n$ here?
– prog_SAHIL
Aug 6 at 14:55
Sorry @MarkViola, fixed
– Cesare
Aug 6 at 14:56
@Cesare Look at the title. Fix that too please.
– Mark Viola
Aug 6 at 14:57
1
Your approach would work if you were looking at the limit as $xtoinfty$. That is where your approach fails here: since $xto1^pm$, we don't have $frac2x+3x^2-1simfrac2xx^2$.
– Clayton
Aug 6 at 15:03
 |Â
show 8 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to find
$$lim_xrightarrow 1^pmleft(frac 2x+3x^2-1right)$$
This is what my textbook does:
$$frac2x+3x^2-1=frac2x+3(x-1)(x+1)=frac 1 x-1cdotfrac2x+3x+1$$
Now, for $xrightarrow1$, we have
$$frac2x+3x+1rightarrow frac 5 2$$
For $xrightarrow1^pm$ we have
$$frac1x-1rightarrowpminfty$$
This means that
$$frac 2x+3x^2-1 rightarrow pminfty$$
This is indeed correct. What I instead did was this:
$$frac 2x+3x^2-1sim frac2xx^2 = frac 2x$$
Then I would "replace" (not sure if this is something I can do) $x$ with $1$ (it's the number which the limit approaches to):
$$frac 2 1 rightarrow2$$
The limit would be $2$. Replacing $x$ with the number that the limit approaches I think is something I learnt in high school. Any hints on why my reasoning is wrong?
limits
edited Aug 6 at 15:13
Daniel Buck
2,3341624
asked Aug 6 at 14:53
Cesare
42129
I want to find
$$lim_xrightarrow 1^pmleft(frac 2x+3x^2-1right)$$
This is what my textbook does:
$$frac2x+3x^2-1=frac2x+3(x-1)(x+1)=frac 1 x-1cdotfrac2x+3x+1$$
Now, for $xrightarrow1$, we have
$$frac2x+3x+1rightarrow frac 5 2$$
For $xrightarrow1^pm$ we have
$$frac1x-1rightarrowpminfty$$
This means that
$$frac 2x+3x^2-1 rightarrow pminfty$$
This is indeed correct. What I instead did was this:
$$frac 2x+3x^2-1sim frac2xx^2 = frac 2x$$
Then I would "replace" (not sure if this is something I can do) $x$ with $1$ (it's the number which the limit approaches to):
$$frac 2 1 rightarrow2$$
The limit would be $2$. Replacing $x$ with the number that the limit approaches I think is something I learnt in high school. Any hints on why my reasoning is wrong?
limits
edited Aug 6 at 15:13
Daniel Buck
2,3341624
asked Aug 6 at 14:53
Cesare
42129
edited Aug 6 at 15:13
Daniel Buck
2,3341624
edited Aug 6 at 15:13
Daniel Buck
2,3341624
edited Aug 6 at 15:13
Daniel Buck
2,3341624
2,3341624
asked Aug 6 at 14:53
Cesare
42129
asked Aug 6 at 14:53
Cesare
42129
asked Aug 6 at 14:53
Cesare
42129
42129
How does $x$ depend on $n$?
– Mark Viola
Aug 6 at 14:55
1
What is $n$ here?
– prog_SAHIL
Aug 6 at 14:55
Sorry @MarkViola, fixed
– Cesare
Aug 6 at 14:56
@Cesare Look at the title. Fix that too please.
– Mark Viola
Aug 6 at 14:57
1
Your approach would work if you were looking at the limit as $xtoinfty$. That is where your approach fails here: since $xto1^pm$, we don't have $frac2x+3x^2-1simfrac2xx^2$.
– Clayton
Aug 6 at 15:03
 |Â
show 8 more comments
How does $x$ depend on $n$?
– Mark Viola
Aug 6 at 14:55
1
What is $n$ here?
– prog_SAHIL
Aug 6 at 14:55
Sorry @MarkViola, fixed
– Cesare
Aug 6 at 14:56
@Cesare Look at the title. Fix that too please.
– Mark Viola
Aug 6 at 14:57
1
Your approach would work if you were looking at the limit as $xtoinfty$. That is where your approach fails here: since $xto1^pm$, we don't have $frac2x+3x^2-1simfrac2xx^2$.
– Clayton
Aug 6 at 15:03
How does $x$ depend on $n$?
– Mark Viola
Aug 6 at 14:55
How does $x$ depend on $n$?
– Mark Viola
Aug 6 at 14:55
1
1
What is $n$ here?
– prog_SAHIL
Aug 6 at 14:55
What is $n$ here?
– prog_SAHIL
Aug 6 at 14:55
Sorry @MarkViola, fixed
– Cesare
Aug 6 at 14:56
Sorry @MarkViola, fixed
– Cesare
Aug 6 at 14:56
@Cesare Look at the title. Fix that too please.
– Mark Viola
Aug 6 at 14:57
@Cesare Look at the title. Fix that too please.
– Mark Viola
Aug 6 at 14:57
1
1
Your approach would work if you were looking at the limit as $xtoinfty$. That is where your approach fails here: since $xto1^pm$, we don't have $frac2x+3x^2-1simfrac2xx^2$.
– Clayton
Aug 6 at 15:03
Your approach would work if you were looking at the limit as $xtoinfty$. That is where your approach fails here: since $xto1^pm$, we don't have $frac2x+3x^2-1simfrac2xx^2$.
– Clayton
Aug 6 at 15:03
 |Â
show 8 more comments
6 Answers
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up vote
1
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accepted
There is no indeterminant form here.
So you can directly evaluate the limits by putting $x=1^+$ and $x=1^-$
Also for your doubt,
You cannot approximate $frac2x+3x^2-1approxfrac2x$. It just makes no sense to do so.
answered Aug 6 at 15:00
prog_SAHIL
864217
add a comment |Â
up vote
2
down vote
The idea $frac2x+3x^2 - 1 sim frac2xx^2$ works only when the limit is to (positive) infinity. This is because we write $frac2x+3x^2-1 = frac2x+32x cdot fracx^2x^2-1 cdot frac2xx^2$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $frac2xx^2$, if the limit were to infinity.
On the other hand, when the limit is as $x to 1^+$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.
On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $frac2x+3x^2 - 1$ gives $frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).
The best way to prove this, is to do what the book does : $frac2x+3x^2-1 = frac1x-1 times frac2x+3x+1$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.
Suppose you really wanted to use an approach using asymptotics. Then, set $y = frac 1x-1$. Then, as $x to 1^+$ we have $y to + infty$.
We get $frac2x+3(x-1)(x+1) = y left(2 + frac1x+1right) = y left(2 + frac y2y + 1right)$.
Now, $fracy2y+1 sim frac 12$, hence $2 + frac y2y+1 sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+infty$.
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
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I think you cannot do $frac2x+3x^2-1 ∼ frac2xx^2$. You can do this only when $x$ is big enough so that finite number can be considered as 0 comparing to $x$. But in your case, $x$ approaches to finite number. Thus, the approximation does not work.
answered Aug 6 at 15:02
Lev Ban
36516
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up vote
1
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Let consider $x=y+1to 0$ with $yto 0$
$$lim_xrightarrow 1^pmfrac 2x+3x^2-1=lim_yrightarrow 0 frac y+5y(y+3)=lim_yrightarrow 0 frac 1y cdot lim_yrightarrow 0 frac y+5y+3=frac53 lim_yrightarrow 0 frac 1y=pm infty$$
answered Aug 6 at 15:05
gimusi
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For $xrightarrow1^+$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow+infty$$
and for $xrightarrow1^-$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow-infty$$
answered Aug 6 at 15:11
Daniel Buck
2,3341624
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up vote
1
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The step
$$
frac 2x+3x^2-1sim frac2xx^2 = frac2x
$$
is completely unjustified and totally wrong.
For $x$ near $1$, $2x+5$ is not near $2x$, nor is $x^2-1$ near $x^2$.
You're confusing this with the limit at $pminfty$, where those steps are possible and meaningful.
answered Aug 6 at 15:48
egreg
164k1180187
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is no indeterminant form here.
So you can directly evaluate the limits by putting $x=1^+$ and $x=1^-$
Also for your doubt,
You cannot approximate $frac2x+3x^2-1approxfrac2x$. It just makes no sense to do so.
answered Aug 6 at 15:00
prog_SAHIL
864217
add a comment |Â
up vote
1
down vote
accepted
There is no indeterminant form here.
So you can directly evaluate the limits by putting $x=1^+$ and $x=1^-$
Also for your doubt,
You cannot approximate $frac2x+3x^2-1approxfrac2x$. It just makes no sense to do so.
answered Aug 6 at 15:00
prog_SAHIL
864217
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is no indeterminant form here.
So you can directly evaluate the limits by putting $x=1^+$ and $x=1^-$
Also for your doubt,
You cannot approximate $frac2x+3x^2-1approxfrac2x$. It just makes no sense to do so.
answered Aug 6 at 15:00
prog_SAHIL
864217
There is no indeterminant form here.
So you can directly evaluate the limits by putting $x=1^+$ and $x=1^-$
Also for your doubt,
You cannot approximate $frac2x+3x^2-1approxfrac2x$. It just makes no sense to do so.
answered Aug 6 at 15:00
prog_SAHIL
864217
answered Aug 6 at 15:00
prog_SAHIL
864217
answered Aug 6 at 15:00
prog_SAHIL
864217
answered Aug 6 at 15:00
prog_SAHIL
864217
864217
add a comment |Â
add a comment |Â
up vote
2
down vote
The idea $frac2x+3x^2 - 1 sim frac2xx^2$ works only when the limit is to (positive) infinity. This is because we write $frac2x+3x^2-1 = frac2x+32x cdot fracx^2x^2-1 cdot frac2xx^2$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $frac2xx^2$, if the limit were to infinity.
On the other hand, when the limit is as $x to 1^+$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.
On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $frac2x+3x^2 - 1$ gives $frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).
The best way to prove this, is to do what the book does : $frac2x+3x^2-1 = frac1x-1 times frac2x+3x+1$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.
Suppose you really wanted to use an approach using asymptotics. Then, set $y = frac 1x-1$. Then, as $x to 1^+$ we have $y to + infty$.
We get $frac2x+3(x-1)(x+1) = y left(2 + frac1x+1right) = y left(2 + frac y2y + 1right)$.
Now, $fracy2y+1 sim frac 12$, hence $2 + frac y2y+1 sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+infty$.
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
add a comment |Â
up vote
2
down vote
The idea $frac2x+3x^2 - 1 sim frac2xx^2$ works only when the limit is to (positive) infinity. This is because we write $frac2x+3x^2-1 = frac2x+32x cdot fracx^2x^2-1 cdot frac2xx^2$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $frac2xx^2$, if the limit were to infinity.
On the other hand, when the limit is as $x to 1^+$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.
On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $frac2x+3x^2 - 1$ gives $frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).
The best way to prove this, is to do what the book does : $frac2x+3x^2-1 = frac1x-1 times frac2x+3x+1$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.
Suppose you really wanted to use an approach using asymptotics. Then, set $y = frac 1x-1$. Then, as $x to 1^+$ we have $y to + infty$.
We get $frac2x+3(x-1)(x+1) = y left(2 + frac1x+1right) = y left(2 + frac y2y + 1right)$.
Now, $fracy2y+1 sim frac 12$, hence $2 + frac y2y+1 sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+infty$.
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The idea $frac2x+3x^2 - 1 sim frac2xx^2$ works only when the limit is to (positive) infinity. This is because we write $frac2x+3x^2-1 = frac2x+32x cdot fracx^2x^2-1 cdot frac2xx^2$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $frac2xx^2$, if the limit were to infinity.
On the other hand, when the limit is as $x to 1^+$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.
On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $frac2x+3x^2 - 1$ gives $frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).
The best way to prove this, is to do what the book does : $frac2x+3x^2-1 = frac1x-1 times frac2x+3x+1$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.
Suppose you really wanted to use an approach using asymptotics. Then, set $y = frac 1x-1$. Then, as $x to 1^+$ we have $y to + infty$.
We get $frac2x+3(x-1)(x+1) = y left(2 + frac1x+1right) = y left(2 + frac y2y + 1right)$.
Now, $fracy2y+1 sim frac 12$, hence $2 + frac y2y+1 sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+infty$.
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
The idea $frac2x+3x^2 - 1 sim frac2xx^2$ works only when the limit is to (positive) infinity. This is because we write $frac2x+3x^2-1 = frac2x+32x cdot fracx^2x^2-1 cdot frac2xx^2$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $frac2xx^2$, if the limit were to infinity.
On the other hand, when the limit is as $x to 1^+$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.
On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $frac2x+3x^2 - 1$ gives $frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).
The best way to prove this, is to do what the book does : $frac2x+3x^2-1 = frac1x-1 times frac2x+3x+1$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.
Suppose you really wanted to use an approach using asymptotics. Then, set $y = frac 1x-1$. Then, as $x to 1^+$ we have $y to + infty$.
We get $frac2x+3(x-1)(x+1) = y left(2 + frac1x+1right) = y left(2 + frac y2y + 1right)$.
Now, $fracy2y+1 sim frac 12$, hence $2 + frac y2y+1 sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+infty$.
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Aug 6 at 15:12
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
32k22463
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add a comment |Â
up vote
1
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I think you cannot do $frac2x+3x^2-1 ∼ frac2xx^2$. You can do this only when $x$ is big enough so that finite number can be considered as 0 comparing to $x$. But in your case, $x$ approaches to finite number. Thus, the approximation does not work.
answered Aug 6 at 15:02
Lev Ban
36516
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up vote
1
down vote
I think you cannot do $frac2x+3x^2-1 ∼ frac2xx^2$. You can do this only when $x$ is big enough so that finite number can be considered as 0 comparing to $x$. But in your case, $x$ approaches to finite number. Thus, the approximation does not work.
answered Aug 6 at 15:02
Lev Ban
36516
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think you cannot do $frac2x+3x^2-1 ∼ frac2xx^2$. You can do this only when $x$ is big enough so that finite number can be considered as 0 comparing to $x$. But in your case, $x$ approaches to finite number. Thus, the approximation does not work.
answered Aug 6 at 15:02
Lev Ban
36516
I think you cannot do $frac2x+3x^2-1 ∼ frac2xx^2$. You can do this only when $x$ is big enough so that finite number can be considered as 0 comparing to $x$. But in your case, $x$ approaches to finite number. Thus, the approximation does not work.
answered Aug 6 at 15:02
Lev Ban
36516
answered Aug 6 at 15:02
Lev Ban
36516
answered Aug 6 at 15:02
Lev Ban
36516
answered Aug 6 at 15:02
Lev Ban
36516
36516
add a comment |Â
add a comment |Â
up vote
1
down vote
Let consider $x=y+1to 0$ with $yto 0$
$$lim_xrightarrow 1^pmfrac 2x+3x^2-1=lim_yrightarrow 0 frac y+5y(y+3)=lim_yrightarrow 0 frac 1y cdot lim_yrightarrow 0 frac y+5y+3=frac53 lim_yrightarrow 0 frac 1y=pm infty$$
answered Aug 6 at 15:05
gimusi
65.4k73684
add a comment |Â
up vote
1
down vote
Let consider $x=y+1to 0$ with $yto 0$
$$lim_xrightarrow 1^pmfrac 2x+3x^2-1=lim_yrightarrow 0 frac y+5y(y+3)=lim_yrightarrow 0 frac 1y cdot lim_yrightarrow 0 frac y+5y+3=frac53 lim_yrightarrow 0 frac 1y=pm infty$$
answered Aug 6 at 15:05
gimusi
65.4k73684
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let consider $x=y+1to 0$ with $yto 0$
$$lim_xrightarrow 1^pmfrac 2x+3x^2-1=lim_yrightarrow 0 frac y+5y(y+3)=lim_yrightarrow 0 frac 1y cdot lim_yrightarrow 0 frac y+5y+3=frac53 lim_yrightarrow 0 frac 1y=pm infty$$
answered Aug 6 at 15:05
gimusi
65.4k73684
Let consider $x=y+1to 0$ with $yto 0$
$$lim_xrightarrow 1^pmfrac 2x+3x^2-1=lim_yrightarrow 0 frac y+5y(y+3)=lim_yrightarrow 0 frac 1y cdot lim_yrightarrow 0 frac y+5y+3=frac53 lim_yrightarrow 0 frac 1y=pm infty$$
answered Aug 6 at 15:05
gimusi
65.4k73684
answered Aug 6 at 15:05
gimusi
65.4k73684
answered Aug 6 at 15:05
gimusi
65.4k73684
answered Aug 6 at 15:05
gimusi
65.4k73684
65.4k73684
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add a comment |Â
up vote
1
down vote
For $xrightarrow1^+$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow+infty$$
and for $xrightarrow1^-$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow-infty$$
answered Aug 6 at 15:11
Daniel Buck
2,3341624
add a comment |Â
up vote
1
down vote
For $xrightarrow1^+$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow+infty$$
and for $xrightarrow1^-$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow-infty$$
answered Aug 6 at 15:11
Daniel Buck
2,3341624
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $xrightarrow1^+$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow+infty$$
and for $xrightarrow1^-$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow-infty$$
answered Aug 6 at 15:11
Daniel Buck
2,3341624
For $xrightarrow1^+$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow+infty$$
and for $xrightarrow1^-$, we have
$$frac2x+3x+1rightarrow frac52quadtextandquadfrac1x-1rightarrow-infty$$
answered Aug 6 at 15:11
Daniel Buck
2,3341624
answered Aug 6 at 15:11
Daniel Buck
2,3341624
answered Aug 6 at 15:11
Daniel Buck
2,3341624
answered Aug 6 at 15:11
Daniel Buck
2,3341624
2,3341624
add a comment |Â
add a comment |Â
up vote
1
down vote
The step
$$
frac 2x+3x^2-1sim frac2xx^2 = frac2x
$$
is completely unjustified and totally wrong.
For $x$ near $1$, $2x+5$ is not near $2x$, nor is $x^2-1$ near $x^2$.
You're confusing this with the limit at $pminfty$, where those steps are possible and meaningful.
answered Aug 6 at 15:48
egreg
164k1180187
add a comment |Â
up vote
1
down vote
The step
$$
frac 2x+3x^2-1sim frac2xx^2 = frac2x
$$
is completely unjustified and totally wrong.
For $x$ near $1$, $2x+5$ is not near $2x$, nor is $x^2-1$ near $x^2$.
You're confusing this with the limit at $pminfty$, where those steps are possible and meaningful.
answered Aug 6 at 15:48
egreg
164k1180187
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The step
$$
frac 2x+3x^2-1sim frac2xx^2 = frac2x
$$
is completely unjustified and totally wrong.
For $x$ near $1$, $2x+5$ is not near $2x$, nor is $x^2-1$ near $x^2$.
You're confusing this with the limit at $pminfty$, where those steps are possible and meaningful.
answered Aug 6 at 15:48
egreg
164k1180187
The step
$$
frac 2x+3x^2-1sim frac2xx^2 = frac2x
$$
is completely unjustified and totally wrong.
For $x$ near $1$, $2x+5$ is not near $2x$, nor is $x^2-1$ near $x^2$.
You're confusing this with the limit at $pminfty$, where those steps are possible and meaningful.
answered Aug 6 at 15:48
egreg
164k1180187
answered Aug 6 at 15:48
egreg
164k1180187
answered Aug 6 at 15:48
egreg
164k1180187
answered Aug 6 at 15:48
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
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How does $x$ depend on $n$?
– Mark Viola
Aug 6 at 14:55
1
What is $n$ here?
– prog_SAHIL
Aug 6 at 14:55
Sorry @MarkViola, fixed
– Cesare
Aug 6 at 14:56
@Cesare Look at the title. Fix that too please.
– Mark Viola
Aug 6 at 14:57
1
Your approach would work if you were looking at the limit as $xtoinfty$. That is where your approach fails here: since $xto1^pm$, we don't have $frac2x+3x^2-1simfrac2xx^2$.
– Clayton
Aug 6 at 15:03