Determine the limit, or show it doesn't exist: $lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$
$begingroup$
Determine the limit of the following or prove it doesn't exist:
$$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$
If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.
calculus limits trigonometry
edited Dec 18 '18 at 3:47
Blue
47.9k870153
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
$endgroup$
add a comment |
$begingroup$
Determine the limit of the following or prove it doesn't exist:
$$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$
If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.
calculus limits trigonometry
edited Dec 18 '18 at 3:47
Blue
47.9k870153
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
$endgroup$
add a comment |
$begingroup$
Determine the limit of the following or prove it doesn't exist:
$$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$
If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.
calculus limits trigonometry
edited Dec 18 '18 at 3:47
Blue
47.9k870153
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
$endgroup$
Determine the limit of the following or prove it doesn't exist:
$$lim_{xto 2} left(arctanleft(frac{1}{2-x}right)right)^2$$
If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.
calculus limits trigonometry
calculus limits trigonometry
edited Dec 18 '18 at 3:47
Blue
47.9k870153
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
edited Dec 18 '18 at 3:47
Blue
47.9k870153
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
edited Dec 18 '18 at 3:47
Blue
47.9k870153
edited Dec 18 '18 at 3:47
Blue
47.9k870153
edited Dec 18 '18 at 3:47
Blue
47.9k870153
47.9k870153
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
asked Dec 18 '18 at 2:50
J. LastinJ. Lastin
986
986
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)
answered Dec 18 '18 at 2:58
william122william122
52412
$endgroup$
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
1
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
add a comment |
$begingroup$
Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.
Just because, let's do it rigorously.
We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.
Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$
Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
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$begingroup$
Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)
answered Dec 18 '18 at 2:58
william122william122
52412
$endgroup$
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
1
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
add a comment |
$begingroup$
Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)
answered Dec 18 '18 at 2:58
william122william122
52412
$endgroup$
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
1
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
add a comment |
$begingroup$
Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)
answered Dec 18 '18 at 2:58
william122william122
52412
$endgroup$
Performing the substitution $u=frac{1}{2-x}$, this is just $limlimits_{utoinfty}(tan^{-1} u)^2$, which evaluates to $frac{pi^2}{4}$, since $limlimits_{utoinfty}tan^{-1} u=frac{pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)
answered Dec 18 '18 at 2:58
william122william122
52412
edited Dec 18 '18 at 3:09
answered Dec 18 '18 at 2:58
william122william122
52412
answered Dec 18 '18 at 2:58
william122william122
52412
answered Dec 18 '18 at 2:58
william122william122
52412
52412
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
1
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
add a comment |
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
1
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
$begingroup$
Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following.
$endgroup$
– J. Lastin
Dec 18 '18 at 3:06
1
1
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
$begingroup$
OK I added a bit
$endgroup$
– william122
Dec 18 '18 at 3:10
add a comment |
$begingroup$
Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.
Just because, let's do it rigorously.
We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.
Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$
Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
$endgroup$
add a comment |
$begingroup$
Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.
Just because, let's do it rigorously.
We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.
Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$
Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
$endgroup$
add a comment |
$begingroup$
Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.
Just because, let's do it rigorously.
We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.
Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$
Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
$endgroup$
Intuitively, as $x$ 2 from the positive side, $frac{1}{2-x}$ approaches $-infty$ and as $x$ approaches 2 from the negative side, $frac{1}{2-x}$ approaches $infty$. However, $arctan$ of $pminfty$ is $pmfrac{pi}{2}$, so squaring this gets $frac{pi^2}{4}$, the value of the limit.
Just because, let's do it rigorously.
We show that the left and right limits exist and that they are equal, taking for granted that $lim_{xtoinfty}arctan(x)=frac{pi}{2}$ and that $lim_{xto -infty}=-frac{pi}{2}$ and that $arctan(x)$ is monotonic.
Consider $lim_{xto2^{-}}arctan^2(x)$. Pick $Y$ s.t. $y>Yrightarrow frac{pi}{2}-arctan(y)<epsilon$ so then $frac{pi^2}{4}-arctan(y)<epsilon(pi-epsilon)$. To get $y>Y$, we need $frac{1}{2-x}>Yrightarrow x>2-frac{1}{Y}$ and, since this is the left limit, $x<2$. So take $delta=2-frac{1}{Y}$ so that $xin(2-delta,2)rightarrow arctan^2(frac{1}{2-x})inBigl(frac{pi^2}{4}-epsilon(pi-epsilon),frac{pi^2}{4}Bigr)$, so $lim_{xto2^-}arctan^2(x)=frac{pi^2}{4}$
Excercise: compute $lim_{xto2^+}arctan^2(x)$ and show they are equal.
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
answered Dec 18 '18 at 3:26
R.JacksonR.Jackson
1688
1688
add a comment |
add a comment |
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