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When do you know when you have to check for positive AND negative limits?
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When I see a simple limit question like "find the limit as $x rightarrow infty$
when $f(x) =$ $3x+7 over x+2$ I know that all I have to do is factor out x which makes $3 over 1$ $cdot$ $7 over x$ / $2 over x$, which both turn into 0 as I plug ONLY positive $infty$ in for x.
But for limits like $e^x over e^x +1$, as $x rightarrow infty$, how do I know that I have to plug in negative $infty$ and positive $infty$ for x after I simplify? (which gives me $1 over 1 + (1/ e^x)$)
Does it have something to do with squeeze theorem? Thanks!
calculus
asked 1 hour ago
ming
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up vote
2
down vote
favorite
When I see a simple limit question like "find the limit as $x rightarrow infty$
when $f(x) =$ $3x+7 over x+2$ I know that all I have to do is factor out x which makes $3 over 1$ $cdot$ $7 over x$ / $2 over x$, which both turn into 0 as I plug ONLY positive $infty$ in for x.
But for limits like $e^x over e^x +1$, as $x rightarrow infty$, how do I know that I have to plug in negative $infty$ and positive $infty$ for x after I simplify? (which gives me $1 over 1 + (1/ e^x)$)
Does it have something to do with squeeze theorem? Thanks!
calculus
asked 1 hour ago
ming
111
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It the limit is $lim x to infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $lim x to -infty$ in which cause you always check negative and never check positive.
– fleablood
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
When I see a simple limit question like "find the limit as $x rightarrow infty$
when $f(x) =$ $3x+7 over x+2$ I know that all I have to do is factor out x which makes $3 over 1$ $cdot$ $7 over x$ / $2 over x$, which both turn into 0 as I plug ONLY positive $infty$ in for x.
But for limits like $e^x over e^x +1$, as $x rightarrow infty$, how do I know that I have to plug in negative $infty$ and positive $infty$ for x after I simplify? (which gives me $1 over 1 + (1/ e^x)$)
Does it have something to do with squeeze theorem? Thanks!
calculus
asked 1 hour ago
ming
111
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
When I see a simple limit question like "find the limit as $x rightarrow infty$
when $f(x) =$ $3x+7 over x+2$ I know that all I have to do is factor out x which makes $3 over 1$ $cdot$ $7 over x$ / $2 over x$, which both turn into 0 as I plug ONLY positive $infty$ in for x.
But for limits like $e^x over e^x +1$, as $x rightarrow infty$, how do I know that I have to plug in negative $infty$ and positive $infty$ for x after I simplify? (which gives me $1 over 1 + (1/ e^x)$)
Does it have something to do with squeeze theorem? Thanks!
calculus
calculus
asked 1 hour ago
ming
111
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
ming
111
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
ming
111
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
ming
111
asked 1 hour ago
ming
111
111
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It the limit is $lim x to infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $lim x to -infty$ in which cause you always check negative and never check positive.
– fleablood
1 hour ago
add a comment |Â
It the limit is $lim x to infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $lim x to -infty$ in which cause you always check negative and never check positive.
– fleablood
1 hour ago
It the limit is $lim x to infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $lim x to -infty$ in which cause you always check negative and never check positive.
– fleablood
1 hour ago
It the limit is $lim x to infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $lim x to -infty$ in which cause you always check negative and never check positive.
– fleablood
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
So a few things to sort out. First of all, no nothing to do with the squeeze theorem. What we can do with the squeeze theorem is say that when we can squeeze a limit between two other limits (or a constant and a limit) whose values we know coincide, then the limit we "squeezed" must agree with that as well.
As far as plugging in $+ infty$ and $-infty$ we never do that. We only take a limit going to a place. $+ infty$ or $-infty$ for example but not both.
What I think you're asking is "how do I know when I have to verify the limit from both the right and the left?" And the answer is, essentially, always. It's just that much of the time our function behaves nicely enough that it's very clear the left and right limits agree. When dealing with limits at $pm infty$ then there is only ever a one sided limit. How do you approach $+ infty$ from the right for example? Well, you can't...
So for your example I'll abuse notation a bit and we have
$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$
answered 1 hour ago
RhythmInk
1,510423
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
add a comment |Â
up vote
1
down vote
If you are told to find $limlimits_xtoinftyf(x)$, then you do not need to consider the case $xto-infty$.
You may be confused between these limits, and finding horizontal asymptotes of a function. To find a horizontal asymptote for a function $f(x)$, you need to check if $limlimits_xtoinftyf(x)$ exists. If it does, it equals some number $L$, and $y=L$ is a horizontal asymptote for $f(x)$. It is prudent in these situations to also to check $limlimits_xto-inftyf(x)$, because some functions might have a horizontal asymptote distinct from the first one we found. $f(x)=tan^-1x$ is an example.
answered 1 hour ago
The Count
2,28361431
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
So a few things to sort out. First of all, no nothing to do with the squeeze theorem. What we can do with the squeeze theorem is say that when we can squeeze a limit between two other limits (or a constant and a limit) whose values we know coincide, then the limit we "squeezed" must agree with that as well.
As far as plugging in $+ infty$ and $-infty$ we never do that. We only take a limit going to a place. $+ infty$ or $-infty$ for example but not both.
What I think you're asking is "how do I know when I have to verify the limit from both the right and the left?" And the answer is, essentially, always. It's just that much of the time our function behaves nicely enough that it's very clear the left and right limits agree. When dealing with limits at $pm infty$ then there is only ever a one sided limit. How do you approach $+ infty$ from the right for example? Well, you can't...
So for your example I'll abuse notation a bit and we have
$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$
answered 1 hour ago
RhythmInk
1,510423
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
add a comment |Â
up vote
3
down vote
So a few things to sort out. First of all, no nothing to do with the squeeze theorem. What we can do with the squeeze theorem is say that when we can squeeze a limit between two other limits (or a constant and a limit) whose values we know coincide, then the limit we "squeezed" must agree with that as well.
As far as plugging in $+ infty$ and $-infty$ we never do that. We only take a limit going to a place. $+ infty$ or $-infty$ for example but not both.
What I think you're asking is "how do I know when I have to verify the limit from both the right and the left?" And the answer is, essentially, always. It's just that much of the time our function behaves nicely enough that it's very clear the left and right limits agree. When dealing with limits at $pm infty$ then there is only ever a one sided limit. How do you approach $+ infty$ from the right for example? Well, you can't...
So for your example I'll abuse notation a bit and we have
$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$
answered 1 hour ago
RhythmInk
1,510423
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
So a few things to sort out. First of all, no nothing to do with the squeeze theorem. What we can do with the squeeze theorem is say that when we can squeeze a limit between two other limits (or a constant and a limit) whose values we know coincide, then the limit we "squeezed" must agree with that as well.
As far as plugging in $+ infty$ and $-infty$ we never do that. We only take a limit going to a place. $+ infty$ or $-infty$ for example but not both.
What I think you're asking is "how do I know when I have to verify the limit from both the right and the left?" And the answer is, essentially, always. It's just that much of the time our function behaves nicely enough that it's very clear the left and right limits agree. When dealing with limits at $pm infty$ then there is only ever a one sided limit. How do you approach $+ infty$ from the right for example? Well, you can't...
So for your example I'll abuse notation a bit and we have
$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$
answered 1 hour ago
RhythmInk
1,510423
So a few things to sort out. First of all, no nothing to do with the squeeze theorem. What we can do with the squeeze theorem is say that when we can squeeze a limit between two other limits (or a constant and a limit) whose values we know coincide, then the limit we "squeezed" must agree with that as well.
As far as plugging in $+ infty$ and $-infty$ we never do that. We only take a limit going to a place. $+ infty$ or $-infty$ for example but not both.
What I think you're asking is "how do I know when I have to verify the limit from both the right and the left?" And the answer is, essentially, always. It's just that much of the time our function behaves nicely enough that it's very clear the left and right limits agree. When dealing with limits at $pm infty$ then there is only ever a one sided limit. How do you approach $+ infty$ from the right for example? Well, you can't...
So for your example I'll abuse notation a bit and we have
$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$
answered 1 hour ago
RhythmInk
1,510423
answered 1 hour ago
RhythmInk
1,510423
answered 1 hour ago
RhythmInk
1,510423
answered 1 hour ago
RhythmInk
1,510423
1,510423
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
add a comment |Â
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
and why does this equal 0 when $x rightarrow - infty$?
– ming
33 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
he explained it rather well at the end, there.
– The Count
29 mins ago
add a comment |Â
up vote
1
down vote
If you are told to find $limlimits_xtoinftyf(x)$, then you do not need to consider the case $xto-infty$.
You may be confused between these limits, and finding horizontal asymptotes of a function. To find a horizontal asymptote for a function $f(x)$, you need to check if $limlimits_xtoinftyf(x)$ exists. If it does, it equals some number $L$, and $y=L$ is a horizontal asymptote for $f(x)$. It is prudent in these situations to also to check $limlimits_xto-inftyf(x)$, because some functions might have a horizontal asymptote distinct from the first one we found. $f(x)=tan^-1x$ is an example.
answered 1 hour ago
The Count
2,28361431
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
add a comment |Â
up vote
1
down vote
If you are told to find $limlimits_xtoinftyf(x)$, then you do not need to consider the case $xto-infty$.
You may be confused between these limits, and finding horizontal asymptotes of a function. To find a horizontal asymptote for a function $f(x)$, you need to check if $limlimits_xtoinftyf(x)$ exists. If it does, it equals some number $L$, and $y=L$ is a horizontal asymptote for $f(x)$. It is prudent in these situations to also to check $limlimits_xto-inftyf(x)$, because some functions might have a horizontal asymptote distinct from the first one we found. $f(x)=tan^-1x$ is an example.
answered 1 hour ago
The Count
2,28361431
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you are told to find $limlimits_xtoinftyf(x)$, then you do not need to consider the case $xto-infty$.
You may be confused between these limits, and finding horizontal asymptotes of a function. To find a horizontal asymptote for a function $f(x)$, you need to check if $limlimits_xtoinftyf(x)$ exists. If it does, it equals some number $L$, and $y=L$ is a horizontal asymptote for $f(x)$. It is prudent in these situations to also to check $limlimits_xto-inftyf(x)$, because some functions might have a horizontal asymptote distinct from the first one we found. $f(x)=tan^-1x$ is an example.
answered 1 hour ago
The Count
2,28361431
If you are told to find $limlimits_xtoinftyf(x)$, then you do not need to consider the case $xto-infty$.
You may be confused between these limits, and finding horizontal asymptotes of a function. To find a horizontal asymptote for a function $f(x)$, you need to check if $limlimits_xtoinftyf(x)$ exists. If it does, it equals some number $L$, and $y=L$ is a horizontal asymptote for $f(x)$. It is prudent in these situations to also to check $limlimits_xto-inftyf(x)$, because some functions might have a horizontal asymptote distinct from the first one we found. $f(x)=tan^-1x$ is an example.
answered 1 hour ago
The Count
2,28361431
answered 1 hour ago
The Count
2,28361431
answered 1 hour ago
The Count
2,28361431
answered 1 hour ago
The Count
2,28361431
2,28361431
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
add a comment |Â
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $infty$ make it 0?
– ming
32 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
30 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
21 mins ago
add a comment |Â
ming is a new contributor. Be nice, and check out our Code of Conduct.
ming is a new contributor. Be nice, and check out our Code of Conduct.
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It the limit is $lim x to infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $lim x to -infty$ in which cause you always check negative and never check positive.
– fleablood
1 hour ago