Mixing
Different results with different methods to same limit of polynomial
$begingroup$
I'm considering the following limit while looking for asymptotes:
$$lim_{xto infty} arctandfrac{sqrt{x^2+1}}{x-1}$$
Going the route I initially tried, I divide both parts of the fraction by $x^4$ which would give $arctan(infty)$, or $frac{pi}{2}$ as the result.
But according to Wolfram, the solution is $dfrac{pi}{4}$, which you get by instead taking $x^2$ out of the square root in the numerator, then dividing both numerator and denominator by $x$ which gives $arctan(1)$, or $frac{pi}{4}$.
My question is why my initial approach didn't yield the right result, and why it isn't allowed to do it like I did.
limits exponentiation radicals limits-without-lhopital
edited Jan 26 at 23:34
Key Flex
8,63761233
asked Jan 26 at 23:13
ArcturusArcturus
1267
$endgroup$
add a comment |
$begingroup$
I'm considering the following limit while looking for asymptotes:
$$lim_{xto infty} arctandfrac{sqrt{x^2+1}}{x-1}$$
Going the route I initially tried, I divide both parts of the fraction by $x^4$ which would give $arctan(infty)$, or $frac{pi}{2}$ as the result.
But according to Wolfram, the solution is $dfrac{pi}{4}$, which you get by instead taking $x^2$ out of the square root in the numerator, then dividing both numerator and denominator by $x$ which gives $arctan(1)$, or $frac{pi}{4}$.
My question is why my initial approach didn't yield the right result, and why it isn't allowed to do it like I did.
limits exponentiation radicals limits-without-lhopital
edited Jan 26 at 23:34
Key Flex
8,63761233
asked Jan 26 at 23:13
ArcturusArcturus
1267
$endgroup$
$begingroup$
Are you trying to evaluate the limit or to find asympotes?
$endgroup$
– K Split X
Jan 26 at 23:19
1
$begingroup$
I don't see how, dividing numerator and denominator by $x^4$, you can obtain $infty$ as a limit for the fraction.
$endgroup$
– Bernard
Jan 26 at 23:23
$begingroup$
@Bernard My logic being taking $x^4$ under the root would make it into $x^2$, and then I would have in the numerator $sqrt{frac{x^2}{x^2}+frac{1}{x^2}} = sqrt{1+frac{1}{x^2}}$. That would leave me with $arctanlim_{xto infty}frac{sqrt{1+frac{1}{x^2}}}{frac{1}{x^3}-frac{1}{x^4}} = arctan(infty)$
$endgroup$
– Arcturus
Jan 26 at 23:46
1
$begingroup$
But if you divide the numerator by $x^4$, it takes $x^8$ under the root, not $x^2$.
$endgroup$
– Bernard
Jan 26 at 23:51
$begingroup$
I'm an utter dunce. I shouldn't practice math this late in the night. Sorry.
$endgroup$
– Arcturus
Jan 27 at 0:01
add a comment |
$begingroup$
I'm considering the following limit while looking for asymptotes:
$$lim_{xto infty} arctandfrac{sqrt{x^2+1}}{x-1}$$
Going the route I initially tried, I divide both parts of the fraction by $x^4$ which would give $arctan(infty)$, or $frac{pi}{2}$ as the result.
But according to Wolfram, the solution is $dfrac{pi}{4}$, which you get by instead taking $x^2$ out of the square root in the numerator, then dividing both numerator and denominator by $x$ which gives $arctan(1)$, or $frac{pi}{4}$.
My question is why my initial approach didn't yield the right result, and why it isn't allowed to do it like I did.
limits exponentiation radicals limits-without-lhopital
edited Jan 26 at 23:34
Key Flex
8,63761233
asked Jan 26 at 23:13
ArcturusArcturus
1267
$endgroup$
I'm considering the following limit while looking for asymptotes:
$$lim_{xto infty} arctandfrac{sqrt{x^2+1}}{x-1}$$
Going the route I initially tried, I divide both parts of the fraction by $x^4$ which would give $arctan(infty)$, or $frac{pi}{2}$ as the result.
But according to Wolfram, the solution is $dfrac{pi}{4}$, which you get by instead taking $x^2$ out of the square root in the numerator, then dividing both numerator and denominator by $x$ which gives $arctan(1)$, or $frac{pi}{4}$.
My question is why my initial approach didn't yield the right result, and why it isn't allowed to do it like I did.
limits exponentiation radicals limits-without-lhopital
limits exponentiation radicals limits-without-lhopital
edited Jan 26 at 23:34
Key Flex
8,63761233
asked Jan 26 at 23:13
ArcturusArcturus
1267
edited Jan 26 at 23:34
Key Flex
8,63761233
asked Jan 26 at 23:13
ArcturusArcturus
1267
edited Jan 26 at 23:34
Key Flex
8,63761233
edited Jan 26 at 23:34
Key Flex
8,63761233
edited Jan 26 at 23:34
Key Flex
8,63761233
8,63761233
asked Jan 26 at 23:13
ArcturusArcturus
1267
asked Jan 26 at 23:13
ArcturusArcturus
1267
asked Jan 26 at 23:13
ArcturusArcturus
1267
1267
$begingroup$
Are you trying to evaluate the limit or to find asympotes?
$endgroup$
– K Split X
Jan 26 at 23:19
1
$begingroup$
I don't see how, dividing numerator and denominator by $x^4$, you can obtain $infty$ as a limit for the fraction.
$endgroup$
– Bernard
Jan 26 at 23:23
$begingroup$
@Bernard My logic being taking $x^4$ under the root would make it into $x^2$, and then I would have in the numerator $sqrt{frac{x^2}{x^2}+frac{1}{x^2}} = sqrt{1+frac{1}{x^2}}$. That would leave me with $arctanlim_{xto infty}frac{sqrt{1+frac{1}{x^2}}}{frac{1}{x^3}-frac{1}{x^4}} = arctan(infty)$
$endgroup$
– Arcturus
Jan 26 at 23:46
1
$begingroup$
But if you divide the numerator by $x^4$, it takes $x^8$ under the root, not $x^2$.
$endgroup$
– Bernard
Jan 26 at 23:51
$begingroup$
I'm an utter dunce. I shouldn't practice math this late in the night. Sorry.
$endgroup$
– Arcturus
Jan 27 at 0:01
add a comment |
$begingroup$
Are you trying to evaluate the limit or to find asympotes?
$endgroup$
– K Split X
Jan 26 at 23:19
1
$begingroup$
I don't see how, dividing numerator and denominator by $x^4$, you can obtain $infty$ as a limit for the fraction.
$endgroup$
– Bernard
Jan 26 at 23:23
$begingroup$
@Bernard My logic being taking $x^4$ under the root would make it into $x^2$, and then I would have in the numerator $sqrt{frac{x^2}{x^2}+frac{1}{x^2}} = sqrt{1+frac{1}{x^2}}$. That would leave me with $arctanlim_{xto infty}frac{sqrt{1+frac{1}{x^2}}}{frac{1}{x^3}-frac{1}{x^4}} = arctan(infty)$
$endgroup$
– Arcturus
Jan 26 at 23:46
1
$begingroup$
But if you divide the numerator by $x^4$, it takes $x^8$ under the root, not $x^2$.
$endgroup$
– Bernard
Jan 26 at 23:51
$begingroup$
I'm an utter dunce. I shouldn't practice math this late in the night. Sorry.
$endgroup$
– Arcturus
Jan 27 at 0:01
$begingroup$
Are you trying to evaluate the limit or to find asympotes?
$endgroup$
– K Split X
Jan 26 at 23:19
$begingroup$
Are you trying to evaluate the limit or to find asympotes?
$endgroup$
– K Split X
Jan 26 at 23:19
1
1
$begingroup$
I don't see how, dividing numerator and denominator by $x^4$, you can obtain $infty$ as a limit for the fraction.
$endgroup$
– Bernard
Jan 26 at 23:23
$begingroup$
I don't see how, dividing numerator and denominator by $x^4$, you can obtain $infty$ as a limit for the fraction.
$endgroup$
– Bernard
Jan 26 at 23:23
$begingroup$
@Bernard My logic being taking $x^4$ under the root would make it into $x^2$, and then I would have in the numerator $sqrt{frac{x^2}{x^2}+frac{1}{x^2}} = sqrt{1+frac{1}{x^2}}$. That would leave me with $arctanlim_{xto infty}frac{sqrt{1+frac{1}{x^2}}}{frac{1}{x^3}-frac{1}{x^4}} = arctan(infty)$
$endgroup$
– Arcturus
Jan 26 at 23:46
$begingroup$
@Bernard My logic being taking $x^4$ under the root would make it into $x^2$, and then I would have in the numerator $sqrt{frac{x^2}{x^2}+frac{1}{x^2}} = sqrt{1+frac{1}{x^2}}$. That would leave me with $arctanlim_{xto infty}frac{sqrt{1+frac{1}{x^2}}}{frac{1}{x^3}-frac{1}{x^4}} = arctan(infty)$
$endgroup$
– Arcturus
Jan 26 at 23:46
1
1
$begingroup$
But if you divide the numerator by $x^4$, it takes $x^8$ under the root, not $x^2$.
$endgroup$
– Bernard
Jan 26 at 23:51
$begingroup$
But if you divide the numerator by $x^4$, it takes $x^8$ under the root, not $x^2$.
$endgroup$
– Bernard
Jan 26 at 23:51
$begingroup$
I'm an utter dunce. I shouldn't practice math this late in the night. Sorry.
$endgroup$
– Arcturus
Jan 27 at 0:01
$begingroup$
I'm an utter dunce. I shouldn't practice math this late in the night. Sorry.
$endgroup$
– Arcturus
Jan 27 at 0:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You rather pull $x^2$ from the square root:
$$
sqrt{x^2+1}=sqrt{x^2left(1+frac{1}{x^2}right)}
=xsqrt{1+frac{1}{x^2}}
$$
(because you can assume $x>0$ as you're evaluating the limit for $xtoinfty$).
Then you similarly pull $x$ from $x-1$ getting
$$
xleft(1-frac{1}{x}right)
$$
then you get
$$
lim_{xtoinfty}arctanfrac{sqrt{1+1/x^2}}{1-1/x}=arctan 1
$$
The method is not that different from yours, but you should beware that $x^4=x^{8/2}=sqrt{x^8}$.
answered Jan 27 at 0:37
egregegreg
184k1486206
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You rather pull $x^2$ from the square root:
$$
sqrt{x^2+1}=sqrt{x^2left(1+frac{1}{x^2}right)}
=xsqrt{1+frac{1}{x^2}}
$$
(because you can assume $x>0$ as you're evaluating the limit for $xtoinfty$).
Then you similarly pull $x$ from $x-1$ getting
$$
xleft(1-frac{1}{x}right)
$$
then you get
$$
lim_{xtoinfty}arctanfrac{sqrt{1+1/x^2}}{1-1/x}=arctan 1
$$
The method is not that different from yours, but you should beware that $x^4=x^{8/2}=sqrt{x^8}$.
answered Jan 27 at 0:37
egregegreg
184k1486206
$endgroup$
add a comment |
$begingroup$
You rather pull $x^2$ from the square root:
$$
sqrt{x^2+1}=sqrt{x^2left(1+frac{1}{x^2}right)}
=xsqrt{1+frac{1}{x^2}}
$$
(because you can assume $x>0$ as you're evaluating the limit for $xtoinfty$).
Then you similarly pull $x$ from $x-1$ getting
$$
xleft(1-frac{1}{x}right)
$$
then you get
$$
lim_{xtoinfty}arctanfrac{sqrt{1+1/x^2}}{1-1/x}=arctan 1
$$
The method is not that different from yours, but you should beware that $x^4=x^{8/2}=sqrt{x^8}$.
answered Jan 27 at 0:37
egregegreg
184k1486206
$endgroup$
add a comment |
$begingroup$
You rather pull $x^2$ from the square root:
$$
sqrt{x^2+1}=sqrt{x^2left(1+frac{1}{x^2}right)}
=xsqrt{1+frac{1}{x^2}}
$$
(because you can assume $x>0$ as you're evaluating the limit for $xtoinfty$).
Then you similarly pull $x$ from $x-1$ getting
$$
xleft(1-frac{1}{x}right)
$$
then you get
$$
lim_{xtoinfty}arctanfrac{sqrt{1+1/x^2}}{1-1/x}=arctan 1
$$
The method is not that different from yours, but you should beware that $x^4=x^{8/2}=sqrt{x^8}$.
answered Jan 27 at 0:37
egregegreg
184k1486206
$endgroup$
You rather pull $x^2$ from the square root:
$$
sqrt{x^2+1}=sqrt{x^2left(1+frac{1}{x^2}right)}
=xsqrt{1+frac{1}{x^2}}
$$
(because you can assume $x>0$ as you're evaluating the limit for $xtoinfty$).
Then you similarly pull $x$ from $x-1$ getting
$$
xleft(1-frac{1}{x}right)
$$
then you get
$$
lim_{xtoinfty}arctanfrac{sqrt{1+1/x^2}}{1-1/x}=arctan 1
$$
The method is not that different from yours, but you should beware that $x^4=x^{8/2}=sqrt{x^8}$.
answered Jan 27 at 0:37
egregegreg
184k1486206
answered Jan 27 at 0:37
egregegreg
184k1486206
answered Jan 27 at 0:37
egregegreg
184k1486206
answered Jan 27 at 0:37
egregegreg
184k1486206
184k1486206
add a comment |
add a comment |
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$begingroup$
Are you trying to evaluate the limit or to find asympotes?
$endgroup$
– K Split X
Jan 26 at 23:19
1
$begingroup$
I don't see how, dividing numerator and denominator by $x^4$, you can obtain $infty$ as a limit for the fraction.
$endgroup$
– Bernard
Jan 26 at 23:23
$begingroup$
@Bernard My logic being taking $x^4$ under the root would make it into $x^2$, and then I would have in the numerator $sqrt{frac{x^2}{x^2}+frac{1}{x^2}} = sqrt{1+frac{1}{x^2}}$. That would leave me with $arctanlim_{xto infty}frac{sqrt{1+frac{1}{x^2}}}{frac{1}{x^3}-frac{1}{x^4}} = arctan(infty)$
$endgroup$
– Arcturus
Jan 26 at 23:46
1
$begingroup$
But if you divide the numerator by $x^4$, it takes $x^8$ under the root, not $x^2$.
$endgroup$
– Bernard
Jan 26 at 23:51
$begingroup$
I'm an utter dunce. I shouldn't practice math this late in the night. Sorry.
$endgroup$
– Arcturus
Jan 27 at 0:01