Evaluate $f(x)=xlog$ for $xrightarrowinfty$
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
I have the following function:
$$f(x)=xlogleft$$
I want to find the limit for $xrightarrow+infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=xlogleft(fracx+23-xright)sim xleft( fracx+23-x-1right)=xleft(frac2x-13-xright)=xleft(frac2x-xright)=-2xrightarrow-infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
limits
asked Aug 20 at 9:39
Cesare
582210
add a comment |Â
up vote
5
down vote
favorite
I have the following function:
$$f(x)=xlogleft$$
I want to find the limit for $xrightarrow+infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=xlogleft(fracx+23-xright)sim xleft( fracx+23-x-1right)=xleft(frac2x-13-xright)=xleft(frac2x-xright)=-2xrightarrow-infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
limits
asked Aug 20 at 9:39
Cesare
582210
You actually can't remove the absolute value since $3-x leqslant 0$ when $x rightarrow infty$ !
– tmaths
Aug 20 at 9:42
Your error comes from the fact that $log usim u-1$ is valid for $utoinfty$, which is not the case of $dfracx+23-x$.
– Bernard
Aug 20 at 9:44
@Bernard my textbook says that if $urightarrow1$, then $logusim u -1$ for $xrightarrow x_0$
– Cesare
Aug 20 at 9:48
@Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value.
– Bernard
Aug 20 at 9:57
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I have the following function:
$$f(x)=xlogleft$$
I want to find the limit for $xrightarrow+infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=xlogleft(fracx+23-xright)sim xleft( fracx+23-x-1right)=xleft(frac2x-13-xright)=xleft(frac2x-xright)=-2xrightarrow-infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
limits
asked Aug 20 at 9:39
Cesare
582210
I have the following function:
$$f(x)=xlogleft$$
I want to find the limit for $xrightarrow+infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=xlogleft(fracx+23-xright)sim xleft( fracx+23-x-1right)=xleft(frac2x-13-xright)=xleft(frac2x-xright)=-2xrightarrow-infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
limits
asked Aug 20 at 9:39
Cesare
582210
asked Aug 20 at 9:39
Cesare
582210
asked Aug 20 at 9:39
Cesare
582210
asked Aug 20 at 9:39
Cesare
582210
582210
You actually can't remove the absolute value since $3-x leqslant 0$ when $x rightarrow infty$ !
– tmaths
Aug 20 at 9:42
Your error comes from the fact that $log usim u-1$ is valid for $utoinfty$, which is not the case of $dfracx+23-x$.
– Bernard
Aug 20 at 9:44
@Bernard my textbook says that if $urightarrow1$, then $logusim u -1$ for $xrightarrow x_0$
– Cesare
Aug 20 at 9:48
@Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value.
– Bernard
Aug 20 at 9:57
add a comment |Â
You actually can't remove the absolute value since $3-x leqslant 0$ when $x rightarrow infty$ !
– tmaths
Aug 20 at 9:42
Your error comes from the fact that $log usim u-1$ is valid for $utoinfty$, which is not the case of $dfracx+23-x$.
– Bernard
Aug 20 at 9:44
@Bernard my textbook says that if $urightarrow1$, then $logusim u -1$ for $xrightarrow x_0$
– Cesare
Aug 20 at 9:48
@Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value.
– Bernard
Aug 20 at 9:57
You actually can't remove the absolute value since $3-x leqslant 0$ when $x rightarrow infty$ !
– tmaths
Aug 20 at 9:42
You actually can't remove the absolute value since $3-x leqslant 0$ when $x rightarrow infty$ !
– tmaths
Aug 20 at 9:42
Your error comes from the fact that $log usim u-1$ is valid for $utoinfty$, which is not the case of $dfracx+23-x$.
– Bernard
Aug 20 at 9:44
Your error comes from the fact that $log usim u-1$ is valid for $utoinfty$, which is not the case of $dfracx+23-x$.
– Bernard
Aug 20 at 9:44
@Bernard my textbook says that if $urightarrow1$, then $logusim u -1$ for $xrightarrow x_0$
– Cesare
Aug 20 at 9:48
@Bernard my textbook says that if $urightarrow1$, then $logusim u -1$ for $xrightarrow x_0$
– Cesare
Aug 20 at 9:48
@Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value.
– Bernard
Aug 20 at 9:57
@Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value.
– Bernard
Aug 20 at 9:57
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
We have that for $x>3$
$$logleft=logleft(fracx+2x-3right)=logleft(1+frac5x-3right)$$
and therefore
$$xlogleft=frac5xx-3fraclogleft(1+frac5x-3right)frac5x-3to 5cdot 1=5$$
answered Aug 20 at 9:42
gimusi
68.7k73685
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
add a comment |Â
up vote
2
down vote
$fracx+23-x = frac53-x - 1 < 0$ when $xto infty$.
Hence $undersetxtoinftytextlimf(x) = undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3 = 5$
where the second to last equality follows from $log(x + 1) = x + o(x) text for xto 0 $ thanks to @gimusi's suggesion.
answered Aug 20 at 9:44
kevin
986
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
maybe you should revise it
– gimusi
Aug 20 at 10:32
add a comment |Â
up vote
2
down vote
Let $x>4$;
$I:= logdfrac = displaystyle int_x-3^x+2(1/t)dt.$
MVT:
$I = (1/r)displaystyle int_x-3^x+2dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r in [x-3,x+2]$.
Hence:
$5dfrac xx+2 le xI le 5dfracxx-3$.
Squeeze :
$lim_x rightarrow infty xI =5$.
answered Aug 20 at 11:59
Peter Szilas
8,0672617
Very nice method Peter!
– gimusi
Aug 20 at 13:36
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
We have that for $x>3$
$$logleft=logleft(fracx+2x-3right)=logleft(1+frac5x-3right)$$
and therefore
$$xlogleft=frac5xx-3fraclogleft(1+frac5x-3right)frac5x-3to 5cdot 1=5$$
answered Aug 20 at 9:42
gimusi
68.7k73685
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
add a comment |Â
up vote
7
down vote
accepted
We have that for $x>3$
$$logleft=logleft(fracx+2x-3right)=logleft(1+frac5x-3right)$$
and therefore
$$xlogleft=frac5xx-3fraclogleft(1+frac5x-3right)frac5x-3to 5cdot 1=5$$
answered Aug 20 at 9:42
gimusi
68.7k73685
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
We have that for $x>3$
$$logleft=logleft(fracx+2x-3right)=logleft(1+frac5x-3right)$$
and therefore
$$xlogleft=frac5xx-3fraclogleft(1+frac5x-3right)frac5x-3to 5cdot 1=5$$
answered Aug 20 at 9:42
gimusi
68.7k73685
We have that for $x>3$
$$logleft=logleft(fracx+2x-3right)=logleft(1+frac5x-3right)$$
and therefore
$$xlogleft=frac5xx-3fraclogleft(1+frac5x-3right)frac5x-3to 5cdot 1=5$$
answered Aug 20 at 9:42
gimusi
68.7k73685
answered Aug 20 at 9:42
gimusi
68.7k73685
answered Aug 20 at 9:42
gimusi
68.7k73685
answered Aug 20 at 9:42
gimusi
68.7k73685
68.7k73685
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
add a comment |Â
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$?
– Cesare
Aug 20 at 9:50
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
@Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye
– gimusi
Aug 20 at 9:54
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
Gimusi. Very nice:))
– Peter Szilas
Aug 20 at 11:05
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
@PeterSzilas Thanks, not so difficult ;)
– gimusi
Aug 20 at 11:44
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
Gimusi.Well, well:))
– Peter Szilas
Aug 20 at 12:04
add a comment |Â
up vote
2
down vote
$fracx+23-x = frac53-x - 1 < 0$ when $xto infty$.
Hence $undersetxtoinftytextlimf(x) = undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3 = 5$
where the second to last equality follows from $log(x + 1) = x + o(x) text for xto 0 $ thanks to @gimusi's suggesion.
answered Aug 20 at 9:44
kevin
986
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
maybe you should revise it
– gimusi
Aug 20 at 10:32
add a comment |Â
up vote
2
down vote
$fracx+23-x = frac53-x - 1 < 0$ when $xto infty$.
Hence $undersetxtoinftytextlimf(x) = undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3 = 5$
where the second to last equality follows from $log(x + 1) = x + o(x) text for xto 0 $ thanks to @gimusi's suggesion.
answered Aug 20 at 9:44
kevin
986
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
maybe you should revise it
– gimusi
Aug 20 at 10:32
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$fracx+23-x = frac53-x - 1 < 0$ when $xto infty$.
Hence $undersetxtoinftytextlimf(x) = undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3 = 5$
where the second to last equality follows from $log(x + 1) = x + o(x) text for xto 0 $ thanks to @gimusi's suggesion.
answered Aug 20 at 9:44
kevin
986
$fracx+23-x = frac53-x - 1 < 0$ when $xto infty$.
Hence $undersetxtoinftytextlimf(x) = undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3 = 5$
where the second to last equality follows from $log(x + 1) = x + o(x) text for xto 0 $ thanks to @gimusi's suggesion.
answered Aug 20 at 9:44
kevin
986
edited Aug 20 at 10:42
answered Aug 20 at 9:44
kevin
986
answered Aug 20 at 9:44
kevin
986
answered Aug 20 at 9:44
kevin
986
986
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
maybe you should revise it
– gimusi
Aug 20 at 10:32
add a comment |Â
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
maybe you should revise it
– gimusi
Aug 20 at 10:32
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
The step $undersetxtoinftytextlim xlog(1 + frac5x - 3) = undersetxtoinftytextlimx frac5x - 3$ is true but we need to justify that, for example by $log(1 + frac5x - 3)=frac5x - 3+o(1/x)$.
– gimusi
Aug 20 at 9:58
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
Yes, I forgot to write it explicitly because I assumed it. Thanks!
– kevin
Aug 20 at 10:00
maybe you should revise it
– gimusi
Aug 20 at 10:32
maybe you should revise it
– gimusi
Aug 20 at 10:32
add a comment |Â
up vote
2
down vote
Let $x>4$;
$I:= logdfrac = displaystyle int_x-3^x+2(1/t)dt.$
MVT:
$I = (1/r)displaystyle int_x-3^x+2dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r in [x-3,x+2]$.
Hence:
$5dfrac xx+2 le xI le 5dfracxx-3$.
Squeeze :
$lim_x rightarrow infty xI =5$.
answered Aug 20 at 11:59
Peter Szilas
8,0672617
Very nice method Peter!
– gimusi
Aug 20 at 13:36
add a comment |Â
up vote
2
down vote
Let $x>4$;
$I:= logdfrac = displaystyle int_x-3^x+2(1/t)dt.$
MVT:
$I = (1/r)displaystyle int_x-3^x+2dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r in [x-3,x+2]$.
Hence:
$5dfrac xx+2 le xI le 5dfracxx-3$.
Squeeze :
$lim_x rightarrow infty xI =5$.
answered Aug 20 at 11:59
Peter Szilas
8,0672617
Very nice method Peter!
– gimusi
Aug 20 at 13:36
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $x>4$;
$I:= logdfrac = displaystyle int_x-3^x+2(1/t)dt.$
MVT:
$I = (1/r)displaystyle int_x-3^x+2dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r in [x-3,x+2]$.
Hence:
$5dfrac xx+2 le xI le 5dfracxx-3$.
Squeeze :
$lim_x rightarrow infty xI =5$.
answered Aug 20 at 11:59
Peter Szilas
8,0672617
Let $x>4$;
$I:= logdfrac = displaystyle int_x-3^x+2(1/t)dt.$
MVT:
$I = (1/r)displaystyle int_x-3^x+2dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r in [x-3,x+2]$.
Hence:
$5dfrac xx+2 le xI le 5dfracxx-3$.
Squeeze :
$lim_x rightarrow infty xI =5$.
answered Aug 20 at 11:59
Peter Szilas
8,0672617
edited Aug 20 at 14:30
answered Aug 20 at 11:59
Peter Szilas
8,0672617
answered Aug 20 at 11:59
Peter Szilas
8,0672617
answered Aug 20 at 11:59
Peter Szilas
8,0672617
8,0672617
Very nice method Peter!
– gimusi
Aug 20 at 13:36
add a comment |Â
Very nice method Peter!
– gimusi
Aug 20 at 13:36
Very nice method Peter!
– gimusi
Aug 20 at 13:36
Very nice method Peter!
– gimusi
Aug 20 at 13:36
add a comment |Â
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You actually can't remove the absolute value since $3-x leqslant 0$ when $x rightarrow infty$ !
– tmaths
Aug 20 at 9:42
Your error comes from the fact that $log usim u-1$ is valid for $utoinfty$, which is not the case of $dfracx+23-x$.
– Bernard
Aug 20 at 9:44
@Bernard my textbook says that if $urightarrow1$, then $logusim u -1$ for $xrightarrow x_0$
– Cesare
Aug 20 at 9:48
@Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value.
– Bernard
Aug 20 at 9:57