calculate limit $lim_{nrightarrowinfty} frac{1}{e^{n^2x}}$, x varying in real numbers

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Consider the calculation of the following limit:
$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}$$



I did these passages... Could you tell me if the whole resolution is formally right?



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}} = frac{lim_{nrightarrowinfty} 1}{lim_{nrightarrowinfty} e^{n^2x}}$$



(This is the passage I'm most uncertain of...)



I try to apply the famous limit: $lim_{nrightarrowinfty} 1^n = 1$, so I substitute 1 with $1^n$



$$= frac{lim_{nrightarrowinfty} 1^n}{lim_{nrightarrowinfty} e^{n^2x}}$$



I do substitute 1 with $1^n$ again



$$frac{lim_{nrightarrowinfty} (1^n)^n}{lim_{nrightarrowinfty} e^{n^2x}} = lim_{nrightarrowinfty} frac{1^{n^2}}{e^{n^2 x}} = lim_{nrightarrowinfty} (frac{1}{e^x})^{n^2}$$



At this point I substitute $n^2 = n'$ and study the base of the power ($frac{1}{e^x}$) at the varying of x, calculating the easy limit $lim_{n'rightarrowinfty} (frac{1}{e^x})^{n'}$for $x<0, x=0$ and $x>0$



It's all correct or there are some errors/imprecisions in the calculation?
I'm very uncertain on the fact that I can, formally, substitute $1^n$ to $1$ only because of $lim_{nrightarrowinfty} 1^n = 1$... Is it possible to substitute one expression with another only because they have the same limit?...






calculus limits







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  • It's not wrong to use the fact that $1^n to 1$, but it's wholly unnecessary. The fact that $frac{1}{e^{n^2x}} = left(frac 1 {e^x}right)^{n^2}$ is just something algebraic.
    – T. Bongers
    Nov 27 at 18:04















up vote
1
down vote

favorite












Consider the calculation of the following limit:
$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}$$



I did these passages... Could you tell me if the whole resolution is formally right?



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}} = frac{lim_{nrightarrowinfty} 1}{lim_{nrightarrowinfty} e^{n^2x}}$$



(This is the passage I'm most uncertain of...)



I try to apply the famous limit: $lim_{nrightarrowinfty} 1^n = 1$, so I substitute 1 with $1^n$



$$= frac{lim_{nrightarrowinfty} 1^n}{lim_{nrightarrowinfty} e^{n^2x}}$$



I do substitute 1 with $1^n$ again



$$frac{lim_{nrightarrowinfty} (1^n)^n}{lim_{nrightarrowinfty} e^{n^2x}} = lim_{nrightarrowinfty} frac{1^{n^2}}{e^{n^2 x}} = lim_{nrightarrowinfty} (frac{1}{e^x})^{n^2}$$



At this point I substitute $n^2 = n'$ and study the base of the power ($frac{1}{e^x}$) at the varying of x, calculating the easy limit $lim_{n'rightarrowinfty} (frac{1}{e^x})^{n'}$for $x<0, x=0$ and $x>0$



It's all correct or there are some errors/imprecisions in the calculation?
I'm very uncertain on the fact that I can, formally, substitute $1^n$ to $1$ only because of $lim_{nrightarrowinfty} 1^n = 1$... Is it possible to substitute one expression with another only because they have the same limit?...






calculus limits







share|cite|improve this question






















  • It's not wrong to use the fact that $1^n to 1$, but it's wholly unnecessary. The fact that $frac{1}{e^{n^2x}} = left(frac 1 {e^x}right)^{n^2}$ is just something algebraic.
    – T. Bongers
    Nov 27 at 18:04













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Consider the calculation of the following limit:
$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}$$



I did these passages... Could you tell me if the whole resolution is formally right?



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}} = frac{lim_{nrightarrowinfty} 1}{lim_{nrightarrowinfty} e^{n^2x}}$$



(This is the passage I'm most uncertain of...)



I try to apply the famous limit: $lim_{nrightarrowinfty} 1^n = 1$, so I substitute 1 with $1^n$



$$= frac{lim_{nrightarrowinfty} 1^n}{lim_{nrightarrowinfty} e^{n^2x}}$$



I do substitute 1 with $1^n$ again



$$frac{lim_{nrightarrowinfty} (1^n)^n}{lim_{nrightarrowinfty} e^{n^2x}} = lim_{nrightarrowinfty} frac{1^{n^2}}{e^{n^2 x}} = lim_{nrightarrowinfty} (frac{1}{e^x})^{n^2}$$



At this point I substitute $n^2 = n'$ and study the base of the power ($frac{1}{e^x}$) at the varying of x, calculating the easy limit $lim_{n'rightarrowinfty} (frac{1}{e^x})^{n'}$for $x<0, x=0$ and $x>0$



It's all correct or there are some errors/imprecisions in the calculation?
I'm very uncertain on the fact that I can, formally, substitute $1^n$ to $1$ only because of $lim_{nrightarrowinfty} 1^n = 1$... Is it possible to substitute one expression with another only because they have the same limit?...






calculus limits







share|cite|improve this question













Consider the calculation of the following limit:
$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}$$



I did these passages... Could you tell me if the whole resolution is formally right?



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}} = frac{lim_{nrightarrowinfty} 1}{lim_{nrightarrowinfty} e^{n^2x}}$$



(This is the passage I'm most uncertain of...)



I try to apply the famous limit: $lim_{nrightarrowinfty} 1^n = 1$, so I substitute 1 with $1^n$



$$= frac{lim_{nrightarrowinfty} 1^n}{lim_{nrightarrowinfty} e^{n^2x}}$$



I do substitute 1 with $1^n$ again



$$frac{lim_{nrightarrowinfty} (1^n)^n}{lim_{nrightarrowinfty} e^{n^2x}} = lim_{nrightarrowinfty} frac{1^{n^2}}{e^{n^2 x}} = lim_{nrightarrowinfty} (frac{1}{e^x})^{n^2}$$



At this point I substitute $n^2 = n'$ and study the base of the power ($frac{1}{e^x}$) at the varying of x, calculating the easy limit $lim_{n'rightarrowinfty} (frac{1}{e^x})^{n'}$for $x<0, x=0$ and $x>0$



It's all correct or there are some errors/imprecisions in the calculation?
I'm very uncertain on the fact that I can, formally, substitute $1^n$ to $1$ only because of $lim_{nrightarrowinfty} 1^n = 1$... Is it possible to substitute one expression with another only because they have the same limit?...






calculus limits




calculus limits






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asked Nov 27 at 18:01









Alessio Martorana

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  • It's not wrong to use the fact that $1^n to 1$, but it's wholly unnecessary. The fact that $frac{1}{e^{n^2x}} = left(frac 1 {e^x}right)^{n^2}$ is just something algebraic.
    – T. Bongers
    Nov 27 at 18:04


















  • It's not wrong to use the fact that $1^n to 1$, but it's wholly unnecessary. The fact that $frac{1}{e^{n^2x}} = left(frac 1 {e^x}right)^{n^2}$ is just something algebraic.
    – T. Bongers
    Nov 27 at 18:04
















It's not wrong to use the fact that $1^n to 1$, but it's wholly unnecessary. The fact that $frac{1}{e^{n^2x}} = left(frac 1 {e^x}right)^{n^2}$ is just something algebraic.
– T. Bongers
Nov 27 at 18:04




It's not wrong to use the fact that $1^n to 1$, but it's wholly unnecessary. The fact that $frac{1}{e^{n^2x}} = left(frac 1 {e^x}right)^{n^2}$ is just something algebraic.
– T. Bongers
Nov 27 at 18:04










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










You're making things more complicated than necessary.



For any $n$, we have $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$. Taking the limit of both sides yields
$$lim_{n to infty} frac{1}{e^{n^2 x}} = lim_{n to infty} left(frac{1}{e^x}right)^{n^2}$$






share|cite|improve this answer





















  • I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
    – Alessio Martorana
    Nov 27 at 22:09




















up vote
-2
down vote













Regarding your main doubt note that $forall n$ and $forall x$ the following identity holds



$$frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}$$



and we don't need to use limit concept for that, it is indeed an algebraic identity, and now passing to the limit we obtain



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}=lim_{nrightarrowinfty}left(frac{1}{e^{x}}right)^{n^2}$$



Then we can simply distinguish the cases




  • $x>0 implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to 0quad$ since $frac{1}{e^{x}}<1$


  • $x=0 implies frac{1}{e^{n^2x}}=1$



  • $x<0implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to inftyquad$ since $frac{1}{e^{x}}>1$


and evaluate the limit for each one.






share|cite|improve this answer























  • The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
    – T. Bongers
    Nov 27 at 18:09










  • @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
    – gimusi
    Nov 27 at 18:17












  • You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
    – T. Bongers
    Nov 27 at 18:19










  • @T.Bongers Ah ok, I add something of specific about that!
    – gimusi
    Nov 27 at 18:21











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










You're making things more complicated than necessary.



For any $n$, we have $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$. Taking the limit of both sides yields
$$lim_{n to infty} frac{1}{e^{n^2 x}} = lim_{n to infty} left(frac{1}{e^x}right)^{n^2}$$






share|cite|improve this answer





















  • I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
    – Alessio Martorana
    Nov 27 at 22:09

















up vote
2
down vote



accepted










You're making things more complicated than necessary.



For any $n$, we have $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$. Taking the limit of both sides yields
$$lim_{n to infty} frac{1}{e^{n^2 x}} = lim_{n to infty} left(frac{1}{e^x}right)^{n^2}$$






share|cite|improve this answer





















  • I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
    – Alessio Martorana
    Nov 27 at 22:09















up vote
2
down vote



accepted







up vote
2
down vote



accepted






You're making things more complicated than necessary.



For any $n$, we have $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$. Taking the limit of both sides yields
$$lim_{n to infty} frac{1}{e^{n^2 x}} = lim_{n to infty} left(frac{1}{e^x}right)^{n^2}$$






share|cite|improve this answer












You're making things more complicated than necessary.



For any $n$, we have $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$. Taking the limit of both sides yields
$$lim_{n to infty} frac{1}{e^{n^2 x}} = lim_{n to infty} left(frac{1}{e^x}right)^{n^2}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 18:05









angryavian

38k23180




38k23180












  • I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
    – Alessio Martorana
    Nov 27 at 22:09




















  • I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
    – Alessio Martorana
    Nov 27 at 22:09


















I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
– Alessio Martorana
Nov 27 at 22:09






I'm pretty fine with the fact that we apply $frac{1}{e^{n^2 x}} = left(frac{1}{e^x}right)^{n^2}$ as an algebraic transformation, but I still don't know if, and eventually when, is possible to substitute one expression with another due to the fact that they have the same limit, like I did substituting $1$ with $1^n$
– Alessio Martorana
Nov 27 at 22:09












up vote
-2
down vote













Regarding your main doubt note that $forall n$ and $forall x$ the following identity holds



$$frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}$$



and we don't need to use limit concept for that, it is indeed an algebraic identity, and now passing to the limit we obtain



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}=lim_{nrightarrowinfty}left(frac{1}{e^{x}}right)^{n^2}$$



Then we can simply distinguish the cases




  • $x>0 implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to 0quad$ since $frac{1}{e^{x}}<1$


  • $x=0 implies frac{1}{e^{n^2x}}=1$



  • $x<0implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to inftyquad$ since $frac{1}{e^{x}}>1$


and evaluate the limit for each one.






share|cite|improve this answer























  • The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
    – T. Bongers
    Nov 27 at 18:09










  • @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
    – gimusi
    Nov 27 at 18:17












  • You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
    – T. Bongers
    Nov 27 at 18:19










  • @T.Bongers Ah ok, I add something of specific about that!
    – gimusi
    Nov 27 at 18:21















up vote
-2
down vote













Regarding your main doubt note that $forall n$ and $forall x$ the following identity holds



$$frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}$$



and we don't need to use limit concept for that, it is indeed an algebraic identity, and now passing to the limit we obtain



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}=lim_{nrightarrowinfty}left(frac{1}{e^{x}}right)^{n^2}$$



Then we can simply distinguish the cases




  • $x>0 implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to 0quad$ since $frac{1}{e^{x}}<1$


  • $x=0 implies frac{1}{e^{n^2x}}=1$



  • $x<0implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to inftyquad$ since $frac{1}{e^{x}}>1$


and evaluate the limit for each one.






share|cite|improve this answer























  • The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
    – T. Bongers
    Nov 27 at 18:09










  • @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
    – gimusi
    Nov 27 at 18:17












  • You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
    – T. Bongers
    Nov 27 at 18:19










  • @T.Bongers Ah ok, I add something of specific about that!
    – gimusi
    Nov 27 at 18:21













up vote
-2
down vote










up vote
-2
down vote









Regarding your main doubt note that $forall n$ and $forall x$ the following identity holds



$$frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}$$



and we don't need to use limit concept for that, it is indeed an algebraic identity, and now passing to the limit we obtain



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}=lim_{nrightarrowinfty}left(frac{1}{e^{x}}right)^{n^2}$$



Then we can simply distinguish the cases




  • $x>0 implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to 0quad$ since $frac{1}{e^{x}}<1$


  • $x=0 implies frac{1}{e^{n^2x}}=1$



  • $x<0implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to inftyquad$ since $frac{1}{e^{x}}>1$


and evaluate the limit for each one.






share|cite|improve this answer














Regarding your main doubt note that $forall n$ and $forall x$ the following identity holds



$$frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}$$



and we don't need to use limit concept for that, it is indeed an algebraic identity, and now passing to the limit we obtain



$$lim_{nrightarrowinfty}frac{1}{e^{n^2x}}=lim_{nrightarrowinfty}left(frac{1}{e^{x}}right)^{n^2}$$



Then we can simply distinguish the cases




  • $x>0 implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to 0quad$ since $frac{1}{e^{x}}<1$


  • $x=0 implies frac{1}{e^{n^2x}}=1$



  • $x<0implies frac{1}{e^{n^2x}}=left(frac{1}{e^{x}}right)^{n^2}to inftyquad$ since $frac{1}{e^{x}}>1$


and evaluate the limit for each one.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 18:24

























answered Nov 27 at 18:04









gimusi

92.3k84495




92.3k84495












  • The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
    – T. Bongers
    Nov 27 at 18:09










  • @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
    – gimusi
    Nov 27 at 18:17












  • You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
    – T. Bongers
    Nov 27 at 18:19










  • @T.Bongers Ah ok, I add something of specific about that!
    – gimusi
    Nov 27 at 18:21


















  • The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
    – T. Bongers
    Nov 27 at 18:09










  • @T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
    – gimusi
    Nov 27 at 18:17












  • You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
    – T. Bongers
    Nov 27 at 18:19










  • @T.Bongers Ah ok, I add something of specific about that!
    – gimusi
    Nov 27 at 18:21
















The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
– T. Bongers
Nov 27 at 18:09




The asker has literally said "calculating the easy limit … for $x < 0, x = 0$ and $x > 0$". Did you read the question and identify the parts of their work they are actually skeptical about?
– T. Bongers
Nov 27 at 18:09












@T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
– gimusi
Nov 27 at 18:17






@T.Bongers Yes you are right. I firstly didn't recognize the doubt properly. Now all should be fixed. Thanks
– gimusi
Nov 27 at 18:17














You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
– T. Bongers
Nov 27 at 18:19




You still haven't made an effort to address the line that says "This is the passage I'm most uncertain of..." You're just repeating a computation that the asker has done without commenting on the validity of how they've rewritten something. Which was the real question.
– T. Bongers
Nov 27 at 18:19












@T.Bongers Ah ok, I add something of specific about that!
– gimusi
Nov 27 at 18:21




@T.Bongers Ah ok, I add something of specific about that!
– gimusi
Nov 27 at 18:21


















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