Mixing
Why is $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} = -1$ and not $1$?
$begingroup$
I've been practicing horizontal asymptotes and I came across a problem that I do not understand.
I understood why $lim_{x rightarrow infty}frac{x - 2}{sqrt{x^2 + 1}} = 1$, but i couldn't understand why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} $ isn't $1$ as well, but $-1$ ?
When I calculated $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} $ in wolfram alpha, the result was $-1$.
If anyone could explain why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} = -1$ or $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} = -1$
I'd appreate it very much!
limits
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
$endgroup$
add a comment |
$begingroup$
I've been practicing horizontal asymptotes and I came across a problem that I do not understand.
I understood why $lim_{x rightarrow infty}frac{x - 2}{sqrt{x^2 + 1}} = 1$, but i couldn't understand why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} $ isn't $1$ as well, but $-1$ ?
When I calculated $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} $ in wolfram alpha, the result was $-1$.
If anyone could explain why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} = -1$ or $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} = -1$
I'd appreate it very much!
limits
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
$endgroup$
1
$begingroup$
$sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative.
$endgroup$
– tilper
Jan 2 at 15:12
$begingroup$
A plot on Wolfram Alpha demonstrates it quickly ...
$endgroup$
– Martin R
Jan 2 at 15:16
add a comment |
$begingroup$
I've been practicing horizontal asymptotes and I came across a problem that I do not understand.
I understood why $lim_{x rightarrow infty}frac{x - 2}{sqrt{x^2 + 1}} = 1$, but i couldn't understand why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} $ isn't $1$ as well, but $-1$ ?
When I calculated $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} $ in wolfram alpha, the result was $-1$.
If anyone could explain why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} = -1$ or $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} = -1$
I'd appreate it very much!
limits
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
$endgroup$
I've been practicing horizontal asymptotes and I came across a problem that I do not understand.
I understood why $lim_{x rightarrow infty}frac{x - 2}{sqrt{x^2 + 1}} = 1$, but i couldn't understand why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} $ isn't $1$ as well, but $-1$ ?
When I calculated $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} $ in wolfram alpha, the result was $-1$.
If anyone could explain why $lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} = -1$ or $lim_{x rightarrow -infty}frac{x}{sqrt{x^2}} = -1$
I'd appreate it very much!
limits
limits
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
asked Jan 2 at 15:10
galaxyworksgalaxyworks
154
154
1
$begingroup$
$sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative.
$endgroup$
– tilper
Jan 2 at 15:12
$begingroup$
A plot on Wolfram Alpha demonstrates it quickly ...
$endgroup$
– Martin R
Jan 2 at 15:16
add a comment |
1
$begingroup$
$sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative.
$endgroup$
– tilper
Jan 2 at 15:12
$begingroup$
A plot on Wolfram Alpha demonstrates it quickly ...
$endgroup$
– Martin R
Jan 2 at 15:16
1
1
$begingroup$
$sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative.
$endgroup$
– tilper
Jan 2 at 15:12
$begingroup$
$sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative.
$endgroup$
– tilper
Jan 2 at 15:12
$begingroup$
A plot on Wolfram Alpha demonstrates it quickly ...
$endgroup$
– Martin R
Jan 2 at 15:16
$begingroup$
A plot on Wolfram Alpha demonstrates it quickly ...
$endgroup$
– Martin R
Jan 2 at 15:16
add a comment |
8 Answers
8
active
oldest
votes
$begingroup$
If $x<0, |x|=-x $ so ${xover{sqrt{x^2}}}$ is ${xover{|x|}}={xover{-x}}=-1$.
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
add a comment |
$begingroup$
Note that
$$sqrt{x^2} = vert xvert$$
so the following holds for negative values of $x$:
$$sqrt{x^2} = -x; quad x < 0$$
Hence, you get
$$frac{x}{-x} = -1$$
answered Jan 2 at 15:13
KM101KM101
5,9361523
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$frac{x}{sqrt{x^2}}=frac{x}{|x|}$$
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Hint:
$$
lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} =
lim_{y rightarrow infty}frac{-y - 2}{sqrt{(-y)^2 + 1}} = -1
$$
answered Jan 2 at 15:12
lhflhf
163k10168390
$endgroup$
add a comment |
$begingroup$
The answer really has nothing to do with limits or asymptotes. The real square root function always returns the nonnegative square root. So for every negative value of $x$
$$
frac{x}{sqrt{x^2}} = frac{x}{|x|} = -1 .
$$
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
The core issue is that $sqrt{x^2}$ does not simplify to $x$ for negative $x$.
The symbol $sqrt{r}$ for a nonnegative real number $r$ denotes the nonnegative root of $r$. This is not altered by the fact that in this case $r=x^2$ for some (negative) $x$.
Generally for a real number $x$ one has that $sqrt{x^2} = |x|$.
Keeping that in mind the issue should resolve itself.
It's also that phenomenon for $sqrt{x^2+1}$. When $x$ tends to $-infty$ this is not close to $x$ but rather close to $|x|$.
answered Jan 2 at 15:20
quid♦quid
36.9k95093
$endgroup$
add a comment |
$begingroup$
Set $y=-x$, where $y>0$(why?).
$-dfrac{y}{sqrt{y^2}} = - dfrac{y}{|y|}=-dfrac{y}{y}=-1.$
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
add a comment |
$begingroup$
Because when $xtoinfty^{+}$ $x$ tends to numbers positives and when $xtoinfty^{-}$ $x$ tends to numbers negatives. If you develop this:
begin{eqnarray}
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{|x|} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{x} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}1 \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& 1 \
end{eqnarray}
And:
begin{eqnarray}
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{|x|} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{-x} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}-1 \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& -1 \
end{eqnarray}
answered Jan 2 at 15:20
El PastaEl Pasta
34515
$endgroup$
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $x<0, |x|=-x $ so ${xover{sqrt{x^2}}}$ is ${xover{|x|}}={xover{-x}}=-1$.
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
add a comment |
$begingroup$
If $x<0, |x|=-x $ so ${xover{sqrt{x^2}}}$ is ${xover{|x|}}={xover{-x}}=-1$.
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
add a comment |
$begingroup$
If $x<0, |x|=-x $ so ${xover{sqrt{x^2}}}$ is ${xover{|x|}}={xover{-x}}=-1$.
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
If $x<0, |x|=-x $ so ${xover{sqrt{x^2}}}$ is ${xover{|x|}}={xover{-x}}=-1$.
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
answered Jan 2 at 15:12
Tsemo AristideTsemo Aristide
56.7k11444
56.7k11444
add a comment |
add a comment |
$begingroup$
Note that
$$sqrt{x^2} = vert xvert$$
so the following holds for negative values of $x$:
$$sqrt{x^2} = -x; quad x < 0$$
Hence, you get
$$frac{x}{-x} = -1$$
answered Jan 2 at 15:13
KM101KM101
5,9361523
$endgroup$
add a comment |
$begingroup$
Note that
$$sqrt{x^2} = vert xvert$$
so the following holds for negative values of $x$:
$$sqrt{x^2} = -x; quad x < 0$$
Hence, you get
$$frac{x}{-x} = -1$$
answered Jan 2 at 15:13
KM101KM101
5,9361523
$endgroup$
add a comment |
$begingroup$
Note that
$$sqrt{x^2} = vert xvert$$
so the following holds for negative values of $x$:
$$sqrt{x^2} = -x; quad x < 0$$
Hence, you get
$$frac{x}{-x} = -1$$
answered Jan 2 at 15:13
KM101KM101
5,9361523
$endgroup$
Note that
$$sqrt{x^2} = vert xvert$$
so the following holds for negative values of $x$:
$$sqrt{x^2} = -x; quad x < 0$$
Hence, you get
$$frac{x}{-x} = -1$$
answered Jan 2 at 15:13
KM101KM101
5,9361523
answered Jan 2 at 15:13
KM101KM101
5,9361523
answered Jan 2 at 15:13
KM101KM101
5,9361523
answered Jan 2 at 15:13
KM101KM101
5,9361523
5,9361523
add a comment |
add a comment |
$begingroup$
Hint: Use that $$frac{x}{sqrt{x^2}}=frac{x}{|x|}$$
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$frac{x}{sqrt{x^2}}=frac{x}{|x|}$$
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$frac{x}{sqrt{x^2}}=frac{x}{|x|}$$
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
Hint: Use that $$frac{x}{sqrt{x^2}}=frac{x}{|x|}$$
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 15:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
add a comment |
add a comment |
$begingroup$
Hint:
$$
lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} =
lim_{y rightarrow infty}frac{-y - 2}{sqrt{(-y)^2 + 1}} = -1
$$
answered Jan 2 at 15:12
lhflhf
163k10168390
$endgroup$
add a comment |
$begingroup$
Hint:
$$
lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} =
lim_{y rightarrow infty}frac{-y - 2}{sqrt{(-y)^2 + 1}} = -1
$$
answered Jan 2 at 15:12
lhflhf
163k10168390
$endgroup$
add a comment |
$begingroup$
Hint:
$$
lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} =
lim_{y rightarrow infty}frac{-y - 2}{sqrt{(-y)^2 + 1}} = -1
$$
answered Jan 2 at 15:12
lhflhf
163k10168390
$endgroup$
Hint:
$$
lim_{x rightarrow -infty}frac{x - 2}{sqrt{x^2 + 1}} =
lim_{y rightarrow infty}frac{-y - 2}{sqrt{(-y)^2 + 1}} = -1
$$
answered Jan 2 at 15:12
lhflhf
163k10168390
answered Jan 2 at 15:12
lhflhf
163k10168390
answered Jan 2 at 15:12
lhflhf
163k10168390
answered Jan 2 at 15:12
lhflhf
163k10168390
163k10168390
add a comment |
add a comment |
$begingroup$
The answer really has nothing to do with limits or asymptotes. The real square root function always returns the nonnegative square root. So for every negative value of $x$
$$
frac{x}{sqrt{x^2}} = frac{x}{|x|} = -1 .
$$
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
The answer really has nothing to do with limits or asymptotes. The real square root function always returns the nonnegative square root. So for every negative value of $x$
$$
frac{x}{sqrt{x^2}} = frac{x}{|x|} = -1 .
$$
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
The answer really has nothing to do with limits or asymptotes. The real square root function always returns the nonnegative square root. So for every negative value of $x$
$$
frac{x}{sqrt{x^2}} = frac{x}{|x|} = -1 .
$$
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
The answer really has nothing to do with limits or asymptotes. The real square root function always returns the nonnegative square root. So for every negative value of $x$
$$
frac{x}{sqrt{x^2}} = frac{x}{|x|} = -1 .
$$
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
answered Jan 2 at 15:14
Ethan BolkerEthan Bolker
42.1k548111
42.1k548111
add a comment |
add a comment |
$begingroup$
The core issue is that $sqrt{x^2}$ does not simplify to $x$ for negative $x$.
The symbol $sqrt{r}$ for a nonnegative real number $r$ denotes the nonnegative root of $r$. This is not altered by the fact that in this case $r=x^2$ for some (negative) $x$.
Generally for a real number $x$ one has that $sqrt{x^2} = |x|$.
Keeping that in mind the issue should resolve itself.
It's also that phenomenon for $sqrt{x^2+1}$. When $x$ tends to $-infty$ this is not close to $x$ but rather close to $|x|$.
answered Jan 2 at 15:20
quid♦quid
36.9k95093
$endgroup$
add a comment |
$begingroup$
The core issue is that $sqrt{x^2}$ does not simplify to $x$ for negative $x$.
The symbol $sqrt{r}$ for a nonnegative real number $r$ denotes the nonnegative root of $r$. This is not altered by the fact that in this case $r=x^2$ for some (negative) $x$.
Generally for a real number $x$ one has that $sqrt{x^2} = |x|$.
Keeping that in mind the issue should resolve itself.
It's also that phenomenon for $sqrt{x^2+1}$. When $x$ tends to $-infty$ this is not close to $x$ but rather close to $|x|$.
answered Jan 2 at 15:20
quid♦quid
36.9k95093
$endgroup$
add a comment |
$begingroup$
The core issue is that $sqrt{x^2}$ does not simplify to $x$ for negative $x$.
The symbol $sqrt{r}$ for a nonnegative real number $r$ denotes the nonnegative root of $r$. This is not altered by the fact that in this case $r=x^2$ for some (negative) $x$.
Generally for a real number $x$ one has that $sqrt{x^2} = |x|$.
Keeping that in mind the issue should resolve itself.
It's also that phenomenon for $sqrt{x^2+1}$. When $x$ tends to $-infty$ this is not close to $x$ but rather close to $|x|$.
answered Jan 2 at 15:20
quid♦quid
36.9k95093
$endgroup$
The core issue is that $sqrt{x^2}$ does not simplify to $x$ for negative $x$.
The symbol $sqrt{r}$ for a nonnegative real number $r$ denotes the nonnegative root of $r$. This is not altered by the fact that in this case $r=x^2$ for some (negative) $x$.
Generally for a real number $x$ one has that $sqrt{x^2} = |x|$.
Keeping that in mind the issue should resolve itself.
It's also that phenomenon for $sqrt{x^2+1}$. When $x$ tends to $-infty$ this is not close to $x$ but rather close to $|x|$.
answered Jan 2 at 15:20
quid♦quid
36.9k95093
answered Jan 2 at 15:20
quid♦quid
36.9k95093
answered Jan 2 at 15:20
quid♦quid
36.9k95093
answered Jan 2 at 15:20
quid♦quid
36.9k95093
36.9k95093
add a comment |
add a comment |
$begingroup$
Set $y=-x$, where $y>0$(why?).
$-dfrac{y}{sqrt{y^2}} = - dfrac{y}{|y|}=-dfrac{y}{y}=-1.$
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
add a comment |
$begingroup$
Set $y=-x$, where $y>0$(why?).
$-dfrac{y}{sqrt{y^2}} = - dfrac{y}{|y|}=-dfrac{y}{y}=-1.$
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
add a comment |
$begingroup$
Set $y=-x$, where $y>0$(why?).
$-dfrac{y}{sqrt{y^2}} = - dfrac{y}{|y|}=-dfrac{y}{y}=-1.$
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
Set $y=-x$, where $y>0$(why?).
$-dfrac{y}{sqrt{y^2}} = - dfrac{y}{|y|}=-dfrac{y}{y}=-1.$
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
answered Jan 2 at 15:22
Peter SzilasPeter Szilas
10.9k2720
10.9k2720
add a comment |
add a comment |
$begingroup$
Because when $xtoinfty^{+}$ $x$ tends to numbers positives and when $xtoinfty^{-}$ $x$ tends to numbers negatives. If you develop this:
begin{eqnarray}
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{|x|} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{x} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}1 \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& 1 \
end{eqnarray}
And:
begin{eqnarray}
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{|x|} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{-x} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}-1 \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& -1 \
end{eqnarray}
answered Jan 2 at 15:20
El PastaEl Pasta
34515
$endgroup$
add a comment |
$begingroup$
Because when $xtoinfty^{+}$ $x$ tends to numbers positives and when $xtoinfty^{-}$ $x$ tends to numbers negatives. If you develop this:
begin{eqnarray}
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{|x|} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{x} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}1 \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& 1 \
end{eqnarray}
And:
begin{eqnarray}
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{|x|} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{-x} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}-1 \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& -1 \
end{eqnarray}
answered Jan 2 at 15:20
El PastaEl Pasta
34515
$endgroup$
add a comment |
$begingroup$
Because when $xtoinfty^{+}$ $x$ tends to numbers positives and when $xtoinfty^{-}$ $x$ tends to numbers negatives. If you develop this:
begin{eqnarray}
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{|x|} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{x} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}1 \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& 1 \
end{eqnarray}
And:
begin{eqnarray}
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{|x|} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{-x} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}-1 \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& -1 \
end{eqnarray}
answered Jan 2 at 15:20
El PastaEl Pasta
34515
$endgroup$
Because when $xtoinfty^{+}$ $x$ tends to numbers positives and when $xtoinfty^{-}$ $x$ tends to numbers negatives. If you develop this:
begin{eqnarray}
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{|x|} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}frac{x}{x} \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{+}}1 \
lim_{xtoinfty^{+}}frac{x}{sqrt{x^2}} &=& 1 \
end{eqnarray}
And:
begin{eqnarray}
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{|x|} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}frac{x}{-x} \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& lim_{xtoinfty^{-}}-1 \
lim_{xtoinfty^{-}}frac{x}{sqrt{x^2}} &=& -1 \
end{eqnarray}
answered Jan 2 at 15:20
El PastaEl Pasta
34515
edited Jan 2 at 15:27
answered Jan 2 at 15:20
El PastaEl Pasta
34515
answered Jan 2 at 15:20
El PastaEl Pasta
34515
answered Jan 2 at 15:20
El PastaEl Pasta
34515
34515
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1
$begingroup$
$sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative.
$endgroup$
– tilper
Jan 2 at 15:12
$begingroup$
A plot on Wolfram Alpha demonstrates it quickly ...
$endgroup$
– Martin R
Jan 2 at 15:16