How to solve the limit $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$
$begingroup$
Next week I have a math exam. While I was doing some exercises I came across this interesting limit:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2})$
After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$
This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?
calculus limits trigonometry infinity
asked Jan 2 at 21:55
MichielvkMichielvk
1065
$endgroup$
add a comment |
$begingroup$
Next week I have a math exam. While I was doing some exercises I came across this interesting limit:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2})$
After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$
This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?
calculus limits trigonometry infinity
asked Jan 2 at 21:55
MichielvkMichielvk
1065
$endgroup$
add a comment |
$begingroup$
Next week I have a math exam. While I was doing some exercises I came across this interesting limit:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2})$
After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$
This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?
calculus limits trigonometry infinity
asked Jan 2 at 21:55
MichielvkMichielvk
1065
$endgroup$
Next week I have a math exam. While I was doing some exercises I came across this interesting limit:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2})$
After struggling a lot, I decided to calculate this limit using my calculator. The answer turns out to be $-1$. The problem is that I don't know how to calculate this limit without a calculator. I tried using L'Hôpital's rule after converting the expression to a fraction. My steps:
$limlimits_{xto infty} (x arctan x - frac{xpi}{2}) = limlimits_{xto infty} frac{2x^2arctan x - x^2pi}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{4xarctan x - frac{2}{x^2+1}-2pi x+2}{2} = limlimits_{xto infty} frac{4x^2arctan x - frac{2x}{x^2+1}-2pi x^2+2x}{2x} stackrel{(H)}{=} limlimits_{xto infty} frac{8xarctan x - frac{2x^2+6}{(x^2+1)^2}-4pi x+6}{2} = dots$
This keeps going on without an end, I also don't see where I can simplify the expression when using L'Hôpital's rule. Am I missing a step or am I using the wrong method? What method can be used instead?
calculus limits trigonometry infinity
calculus limits trigonometry infinity
asked Jan 2 at 21:55
MichielvkMichielvk
1065
asked Jan 2 at 21:55
MichielvkMichielvk
1065
asked Jan 2 at 21:55
MichielvkMichielvk
1065
asked Jan 2 at 21:55
MichielvkMichielvk
1065
asked Jan 2 at 21:55
MichielvkMichielvk
1065
1065
add a comment |
add a comment |
3 Answers
3
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oldest
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Observe
begin{align}
lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
end{align}
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
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add a comment |
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If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
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add a comment |
$begingroup$
You can do the substitution $x=1/t$, recalling that, for $t>0$,
$$
arctanfrac{1}{t}=frac{pi}{2}-arctan t
$$
Thus the limit becomes
$$
lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
$$
Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
$$
lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
$$
This is the reciprocal of the derivative at $pi/2$ of the cotangent.
answered Jan 2 at 22:41
egregegreg
185k1486206
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe
begin{align}
lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
end{align}
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
$endgroup$
add a comment |
$begingroup$
Observe
begin{align}
lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
end{align}
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
$endgroup$
add a comment |
$begingroup$
Observe
begin{align}
lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
end{align}
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
$endgroup$
Observe
begin{align}
lim_{xrightarrow infty} xarctan x-xfrac{pi}{2}=lim_{xrightarrowinfty}frac{arctan x-frac{pi}{2}}{x^{-1}} = lim_{xrightarrow infty} frac{frac{1}{1+x^2}}{-x^{-2}}=-1
end{align}
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
answered Jan 2 at 22:01
Jacky ChongJacky Chong
19.3k21129
19.3k21129
add a comment |
add a comment |
$begingroup$
If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
$endgroup$
add a comment |
$begingroup$
If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
$endgroup$
add a comment |
$begingroup$
If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
$endgroup$
If you know that $lim_{xto+infty}arctan x = pi/2$, then factoring out $x$ in your original limit you would get a $+infty cdot 0$. You want to apply L'Hospital's Rule. Motivated by this, we do $$lim_{xto +infty} left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} xleft(arctan x - frac{pi}{2}right) = lim_{x to infty} frac{arctan x - (pi/2)}{1/x}.$$Now apply L'Hospital to get $$lim_{x to +infty}left(xarctan x - frac{xpi}{2}right) = lim_{xto +infty} frac{1/(1+x^2)}{-1/x^2} = -lim_{x to +infty}frac{x^2}{1+x^2} = -1.$$
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
answered Jan 2 at 22:03
Ivo TerekIvo Terek
46.6k954146
46.6k954146
add a comment |
add a comment |
$begingroup$
You can do the substitution $x=1/t$, recalling that, for $t>0$,
$$
arctanfrac{1}{t}=frac{pi}{2}-arctan t
$$
Thus the limit becomes
$$
lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
$$
Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
$$
lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
$$
This is the reciprocal of the derivative at $pi/2$ of the cotangent.
answered Jan 2 at 22:41
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
You can do the substitution $x=1/t$, recalling that, for $t>0$,
$$
arctanfrac{1}{t}=frac{pi}{2}-arctan t
$$
Thus the limit becomes
$$
lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
$$
Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
$$
lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
$$
This is the reciprocal of the derivative at $pi/2$ of the cotangent.
answered Jan 2 at 22:41
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
You can do the substitution $x=1/t$, recalling that, for $t>0$,
$$
arctanfrac{1}{t}=frac{pi}{2}-arctan t
$$
Thus the limit becomes
$$
lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
$$
Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
$$
lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
$$
This is the reciprocal of the derivative at $pi/2$ of the cotangent.
answered Jan 2 at 22:41
egregegreg
185k1486206
$endgroup$
You can do the substitution $x=1/t$, recalling that, for $t>0$,
$$
arctanfrac{1}{t}=frac{pi}{2}-arctan t
$$
Thus the limit becomes
$$
lim_{tto0^+}left(frac{1}{t}frac{pi}{2}-frac{1}{t}arctan t-frac{1}{t}frac{pi}{2}right)=lim_{tto0^+}-frac{arctan t}{t}
$$
Alternatively, substitute $u=arctan x$, so $x=tan u$ and the limit becomes
$$
lim_{utopi/2^-}(u-pi/2)tan u=lim_{utopi/2^-}frac{u-pi/2}{cot u}
$$
This is the reciprocal of the derivative at $pi/2$ of the cotangent.
answered Jan 2 at 22:41
egregegreg
185k1486206
answered Jan 2 at 22:41
egregegreg
185k1486206
answered Jan 2 at 22:41
egregegreg
185k1486206
answered Jan 2 at 22:41
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
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