Limit $lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$
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I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$
limits
New contributor
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up vote
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favorite
I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$
limits
New contributor
Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48
I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50
1
Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02
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0
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favorite
up vote
0
down vote
favorite
I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$
limits
New contributor
I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$
limits
limits
New contributor
New contributor
edited Dec 1 at 13:00
Arjang
5,55162363
5,55162363
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asked Dec 1 at 12:45
Michael
23
23
New contributor
New contributor
Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48
I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50
1
Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02
add a comment |
Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48
I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50
1
Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02
Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48
Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48
I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50
I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50
1
1
Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02
Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02
add a comment |
7 Answers
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up vote
1
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accepted
Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
add a comment |
up vote
1
down vote
Since $$log(1+t) sim t$$ as $t to 0$, we have that
$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$
add a comment |
up vote
1
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HINT
$$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$
add a comment |
up vote
1
down vote
As an alternative way, you can also note that
$$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$
$$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$
and
$$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$
add a comment |
up vote
1
down vote
First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.
Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$
2
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
add a comment |
up vote
1
down vote
$lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$
$log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$
$log left(dfrac{e^7}{e^2}right)= 5.$
Used :
$lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.
add a comment |
up vote
0
down vote
$displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.
$$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$
Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.
Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!
add a comment |
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
add a comment |
up vote
1
down vote
accepted
Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$
Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$
answered Dec 1 at 12:55
Bernard
116k637108
116k637108
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
add a comment |
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
– gimusi
Dec 1 at 13:47
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
– gimusi
Dec 1 at 13:49
add a comment |
up vote
1
down vote
Since $$log(1+t) sim t$$ as $t to 0$, we have that
$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$
add a comment |
up vote
1
down vote
Since $$log(1+t) sim t$$ as $t to 0$, we have that
$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $$log(1+t) sim t$$ as $t to 0$, we have that
$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$
Since $$log(1+t) sim t$$ as $t to 0$, we have that
$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$
answered Dec 1 at 12:54
Ant
17.3k22873
17.3k22873
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up vote
1
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HINT
$$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$
add a comment |
up vote
1
down vote
HINT
$$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT
$$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$
HINT
$$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$
answered Dec 1 at 12:56
gimusi
89.7k74495
89.7k74495
add a comment |
add a comment |
up vote
1
down vote
As an alternative way, you can also note that
$$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$
$$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$
and
$$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$
add a comment |
up vote
1
down vote
As an alternative way, you can also note that
$$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$
$$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$
and
$$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$
add a comment |
up vote
1
down vote
up vote
1
down vote
As an alternative way, you can also note that
$$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$
$$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$
and
$$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$
As an alternative way, you can also note that
$$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$
$$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$
and
$$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$
answered Dec 1 at 13:09
KM101
3,273417
3,273417
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add a comment |
up vote
1
down vote
First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.
Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$
2
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
add a comment |
up vote
1
down vote
First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.
Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$
2
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
add a comment |
up vote
1
down vote
up vote
1
down vote
First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.
Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$
First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.
Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$
edited Dec 1 at 13:18
answered Dec 1 at 12:56
Sujit Bhattacharyya
940316
940316
2
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
add a comment |
2
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
2
2
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
– Botond
Dec 1 at 13:09
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
Thanks for pointing that. I edited.
– Sujit Bhattacharyya
Dec 1 at 13:18
add a comment |
up vote
1
down vote
$lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$
$log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$
$log left(dfrac{e^7}{e^2}right)= 5.$
Used :
$lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.
add a comment |
up vote
1
down vote
$lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$
$log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$
$log left(dfrac{e^7}{e^2}right)= 5.$
Used :
$lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.
add a comment |
up vote
1
down vote
up vote
1
down vote
$lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$
$log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$
$log left(dfrac{e^7}{e^2}right)= 5.$
Used :
$lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.
$lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$
$log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$
$log left(dfrac{e^7}{e^2}right)= 5.$
Used :
$lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.
edited Dec 1 at 13:45
answered Dec 1 at 13:39
Peter Szilas
10.4k2720
10.4k2720
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$displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.
$$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$
Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.
Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!
add a comment |
up vote
0
down vote
$displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.
$$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$
Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.
Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!
add a comment |
up vote
0
down vote
up vote
0
down vote
$displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.
$$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$
Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.
Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!
$displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.
$$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$
Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.
Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!
answered yesterday
Paras Khosla
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449
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Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48
I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50
1
Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02