How to find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$...
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As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?
I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.
Thank you for your time.
EDIT:
Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?
limits
asked Oct 9 at 21:25
Edoardo Serra
32
add a comment |
up vote
0
down vote
favorite
As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?
I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.
Thank you for your time.
EDIT:
Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?
limits
asked Oct 9 at 21:25
Edoardo Serra
32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?
I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.
Thank you for your time.
EDIT:
Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?
limits
asked Oct 9 at 21:25
Edoardo Serra
32
As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?
I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.
Thank you for your time.
EDIT:
Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?
limits
limits
asked Oct 9 at 21:25
Edoardo Serra
32
asked Oct 9 at 21:25
Edoardo Serra
32
edited Nov 20 at 18:08
asked Oct 9 at 21:25
Edoardo Serra
32
asked Oct 9 at 21:25
Edoardo Serra
32
asked Oct 9 at 21:25
Edoardo Serra
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
By inspection,
$$
f(x)=frac{i}{sqrt{1-x^2}}
$$
does the job.
The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.
answered Oct 9 at 21:30
qbert
21.9k32459
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
1
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By inspection,
$$
f(x)=frac{i}{sqrt{1-x^2}}
$$
does the job.
The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.
answered Oct 9 at 21:30
qbert
21.9k32459
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
1
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
add a comment |
up vote
0
down vote
accepted
By inspection,
$$
f(x)=frac{i}{sqrt{1-x^2}}
$$
does the job.
The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.
answered Oct 9 at 21:30
qbert
21.9k32459
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
1
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By inspection,
$$
f(x)=frac{i}{sqrt{1-x^2}}
$$
does the job.
The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.
answered Oct 9 at 21:30
qbert
21.9k32459
By inspection,
$$
f(x)=frac{i}{sqrt{1-x^2}}
$$
does the job.
The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.
answered Oct 9 at 21:30
qbert
21.9k32459
answered Oct 9 at 21:30
qbert
21.9k32459
answered Oct 9 at 21:30
qbert
21.9k32459
answered Oct 9 at 21:30
qbert
21.9k32459
21.9k32459
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
1
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
add a comment |
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
1
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
– Edoardo Serra
Oct 9 at 21:34
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
– qbert
Oct 9 at 21:36
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
– Edoardo Serra
Oct 9 at 21:43
1
1
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
sorry, it works, thanks a lot. I've misread the function
– Edoardo Serra
Oct 9 at 21:44
add a comment |
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