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!










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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














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!










share|cite|improve this question







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












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!










share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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asked 1 hour ago









ming

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ming is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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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
















  • 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










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



$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$






share|cite|improve this answer




















  • and why does this equal 0 when $x rightarrow - infty$?
    – ming
    35 mins ago










  • he explained it rather well at the end, there.
    – The Count
    30 mins ago

















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.






share|cite|improve this answer




















  • 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
    33 mins ago










  • As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
    – The Count
    32 mins ago











  • but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
    – ming
    23 mins ago











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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



$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$






share|cite|improve this answer




















  • and why does this equal 0 when $x rightarrow - infty$?
    – ming
    35 mins ago










  • he explained it rather well at the end, there.
    – The Count
    30 mins ago














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



$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$






share|cite|improve this answer




















  • and why does this equal 0 when $x rightarrow - infty$?
    – ming
    35 mins ago










  • he explained it rather well at the end, there.
    – The Count
    30 mins ago












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



$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$






share|cite|improve this answer












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



$$lim_X to infty frace^xe^x+1=lim_x to infty frac11+frac1e^x=frac11+frac1e^infty=frac11+frac1infty=frac11+0=1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









RhythmInk

1,510423




1,510423











  • and why does this equal 0 when $x rightarrow - infty$?
    – ming
    35 mins ago










  • he explained it rather well at the end, there.
    – The Count
    30 mins ago
















  • and why does this equal 0 when $x rightarrow - infty$?
    – ming
    35 mins ago










  • he explained it rather well at the end, there.
    – The Count
    30 mins ago















and why does this equal 0 when $x rightarrow - infty$?
– ming
35 mins ago




and why does this equal 0 when $x rightarrow - infty$?
– ming
35 mins ago












he explained it rather well at the end, there.
– The Count
30 mins ago




he explained it rather well at the end, there.
– The Count
30 mins ago










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.






share|cite|improve this answer




















  • 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
    33 mins ago










  • As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
    – The Count
    32 mins ago











  • but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
    – ming
    23 mins ago















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.






share|cite|improve this answer




















  • 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
    33 mins ago










  • As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
    – The Count
    32 mins ago











  • but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
    – ming
    23 mins ago













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
    33 mins ago










  • As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
    – The Count
    32 mins ago











  • but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
    – ming
    23 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
    33 mins ago










  • As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
    – The Count
    32 mins ago











  • but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
    – ming
    23 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
33 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
33 mins ago












As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
32 mins ago





As $xtoinfty$, then $e^xtoinfty$, but as $xto-infty$, $e^xto 0$.
– The Count
32 mins ago













but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
23 mins ago





but if $e^x rightarrow 0$ then wouldn't $1 over 1 + 1/0$ be the case, which is undefined?
– ming
23 mins ago











ming is a new contributor. Be nice, and check out our Code of Conduct.









 

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