Mixing
$ lim_{xto 0} left(frac{5}{2+sqrt{9+x}}right)^{operatorname{cosec} x}$
$begingroup$
I solved this question in two different ways and got two different answers to it.
Method #1:
The question is of the form $1^infty$.
So, to solve it, I took it in the power of e and wrote the numerator in the natural logarithmic function.
Now, according to the Maclauren series, $$lim_{xto 0} left(frac{ln (x+1)}{x}right) = 1$$
Using this result, then rationalising the power and then using the result $$lim_{xto 0} left(frac{sin x}{x}right) = 1$$
we get the answer as $e^{-1/30}$
Method #2:
In this method, I decided to go by the basics instead.
I took out the left-hand limit and the right-hand limit separately.
If we want the right-hand limit, then we will have to put $x = 0^+$.
Upon doing so, the denominator becomes slightly greater than the numerator, and we can say that the part inside the parenthesis is slightly less than 1.
Now $cosec 0^+$ tends to $infty$
Also, a number less than 1 raised to the power of infinity tends to 0.
$therefore$ The right-hand limit is zero.
In a similar manner when we take out the left-hand limit, that also comes out to be zero.
Hence, by this method, the answer should be 0.
Please help.
limits limits-without-lhopital
edited Jan 22 at 17:15
Fly by Night
26k32978
asked Jan 22 at 16:45
PranayPranay
84
$endgroup$
add a comment |
$begingroup$
I solved this question in two different ways and got two different answers to it.
Method #1:
The question is of the form $1^infty$.
So, to solve it, I took it in the power of e and wrote the numerator in the natural logarithmic function.
Now, according to the Maclauren series, $$lim_{xto 0} left(frac{ln (x+1)}{x}right) = 1$$
Using this result, then rationalising the power and then using the result $$lim_{xto 0} left(frac{sin x}{x}right) = 1$$
we get the answer as $e^{-1/30}$
Method #2:
In this method, I decided to go by the basics instead.
I took out the left-hand limit and the right-hand limit separately.
If we want the right-hand limit, then we will have to put $x = 0^+$.
Upon doing so, the denominator becomes slightly greater than the numerator, and we can say that the part inside the parenthesis is slightly less than 1.
Now $cosec 0^+$ tends to $infty$
Also, a number less than 1 raised to the power of infinity tends to 0.
$therefore$ The right-hand limit is zero.
In a similar manner when we take out the left-hand limit, that also comes out to be zero.
Hence, by this method, the answer should be 0.
Please help.
limits limits-without-lhopital
edited Jan 22 at 17:15
Fly by Night
26k32978
asked Jan 22 at 16:45
PranayPranay
84
$endgroup$
$begingroup$
Welcome to Math StackExchange. While your question is good, please change your question title to describe your doubt better.
$endgroup$
– Swapnil Rustagi
Jan 22 at 16:54
$begingroup$
Please consider using left( and right) to get brackets that match the height of what's between them: $$left(frac{5}{x^2}right)$$ instead of $$(frac{5}{x^2})$$
$endgroup$
– Fly by Night
Jan 22 at 17:17
1
$begingroup$
"a number less than 1 raised to the power of infinity tends to 0" This is where you go wrong. Consider that, for example, $$limleft(1-frac4nright)^n=e^{-4}$$ but $$limleft(1-frac4{sqrt n}right)^n=0$$ and $$limleft(1-frac4{n^2}right)^n=1$$ although $$limleft(1-frac4nright)=limleft(1-frac4{sqrt n}right)=limleft(1-frac4{n^2}right)=1$$
$endgroup$
– Did
Jan 22 at 17:20
add a comment |
$begingroup$
I solved this question in two different ways and got two different answers to it.
Method #1:
The question is of the form $1^infty$.
So, to solve it, I took it in the power of e and wrote the numerator in the natural logarithmic function.
Now, according to the Maclauren series, $$lim_{xto 0} left(frac{ln (x+1)}{x}right) = 1$$
Using this result, then rationalising the power and then using the result $$lim_{xto 0} left(frac{sin x}{x}right) = 1$$
we get the answer as $e^{-1/30}$
Method #2:
In this method, I decided to go by the basics instead.
I took out the left-hand limit and the right-hand limit separately.
If we want the right-hand limit, then we will have to put $x = 0^+$.
Upon doing so, the denominator becomes slightly greater than the numerator, and we can say that the part inside the parenthesis is slightly less than 1.
Now $cosec 0^+$ tends to $infty$
Also, a number less than 1 raised to the power of infinity tends to 0.
$therefore$ The right-hand limit is zero.
In a similar manner when we take out the left-hand limit, that also comes out to be zero.
Hence, by this method, the answer should be 0.
Please help.
limits limits-without-lhopital
edited Jan 22 at 17:15
Fly by Night
26k32978
asked Jan 22 at 16:45
PranayPranay
84
$endgroup$
I solved this question in two different ways and got two different answers to it.
Method #1:
The question is of the form $1^infty$.
So, to solve it, I took it in the power of e and wrote the numerator in the natural logarithmic function.
Now, according to the Maclauren series, $$lim_{xto 0} left(frac{ln (x+1)}{x}right) = 1$$
Using this result, then rationalising the power and then using the result $$lim_{xto 0} left(frac{sin x}{x}right) = 1$$
we get the answer as $e^{-1/30}$
Method #2:
In this method, I decided to go by the basics instead.
I took out the left-hand limit and the right-hand limit separately.
If we want the right-hand limit, then we will have to put $x = 0^+$.
Upon doing so, the denominator becomes slightly greater than the numerator, and we can say that the part inside the parenthesis is slightly less than 1.
Now $cosec 0^+$ tends to $infty$
Also, a number less than 1 raised to the power of infinity tends to 0.
$therefore$ The right-hand limit is zero.
In a similar manner when we take out the left-hand limit, that also comes out to be zero.
Hence, by this method, the answer should be 0.
Please help.
limits limits-without-lhopital
limits limits-without-lhopital
edited Jan 22 at 17:15
Fly by Night
26k32978
asked Jan 22 at 16:45
PranayPranay
84
edited Jan 22 at 17:15
Fly by Night
26k32978
asked Jan 22 at 16:45
PranayPranay
84
edited Jan 22 at 17:15
Fly by Night
26k32978
edited Jan 22 at 17:15
Fly by Night
26k32978
edited Jan 22 at 17:15
Fly by Night
26k32978
26k32978
asked Jan 22 at 16:45
PranayPranay
84
asked Jan 22 at 16:45
PranayPranay
84
asked Jan 22 at 16:45
PranayPranay
84
84
$begingroup$
Welcome to Math StackExchange. While your question is good, please change your question title to describe your doubt better.
$endgroup$
– Swapnil Rustagi
Jan 22 at 16:54
$begingroup$
Please consider using left( and right) to get brackets that match the height of what's between them: $$left(frac{5}{x^2}right)$$ instead of $$(frac{5}{x^2})$$
$endgroup$
– Fly by Night
Jan 22 at 17:17
1
$begingroup$
"a number less than 1 raised to the power of infinity tends to 0" This is where you go wrong. Consider that, for example, $$limleft(1-frac4nright)^n=e^{-4}$$ but $$limleft(1-frac4{sqrt n}right)^n=0$$ and $$limleft(1-frac4{n^2}right)^n=1$$ although $$limleft(1-frac4nright)=limleft(1-frac4{sqrt n}right)=limleft(1-frac4{n^2}right)=1$$
$endgroup$
– Did
Jan 22 at 17:20
add a comment |
$begingroup$
Welcome to Math StackExchange. While your question is good, please change your question title to describe your doubt better.
$endgroup$
– Swapnil Rustagi
Jan 22 at 16:54
$begingroup$
Please consider using left( and right) to get brackets that match the height of what's between them: $$left(frac{5}{x^2}right)$$ instead of $$(frac{5}{x^2})$$
$endgroup$
– Fly by Night
Jan 22 at 17:17
1
$begingroup$
"a number less than 1 raised to the power of infinity tends to 0" This is where you go wrong. Consider that, for example, $$limleft(1-frac4nright)^n=e^{-4}$$ but $$limleft(1-frac4{sqrt n}right)^n=0$$ and $$limleft(1-frac4{n^2}right)^n=1$$ although $$limleft(1-frac4nright)=limleft(1-frac4{sqrt n}right)=limleft(1-frac4{n^2}right)=1$$
$endgroup$
– Did
Jan 22 at 17:20
$begingroup$
Welcome to Math StackExchange. While your question is good, please change your question title to describe your doubt better.
$endgroup$
– Swapnil Rustagi
Jan 22 at 16:54
$begingroup$
Welcome to Math StackExchange. While your question is good, please change your question title to describe your doubt better.
$endgroup$
– Swapnil Rustagi
Jan 22 at 16:54
$begingroup$
Please consider using left( and right) to get brackets that match the height of what's between them: $$left(frac{5}{x^2}right)$$ instead of $$(frac{5}{x^2})$$
$endgroup$
– Fly by Night
Jan 22 at 17:17
$begingroup$
Please consider using left( and right) to get brackets that match the height of what's between them: $$left(frac{5}{x^2}right)$$ instead of $$(frac{5}{x^2})$$
$endgroup$
– Fly by Night
Jan 22 at 17:17
1
1
$begingroup$
"a number less than 1 raised to the power of infinity tends to 0" This is where you go wrong. Consider that, for example, $$limleft(1-frac4nright)^n=e^{-4}$$ but $$limleft(1-frac4{sqrt n}right)^n=0$$ and $$limleft(1-frac4{n^2}right)^n=1$$ although $$limleft(1-frac4nright)=limleft(1-frac4{sqrt n}right)=limleft(1-frac4{n^2}right)=1$$
$endgroup$
– Did
Jan 22 at 17:20
$begingroup$
"a number less than 1 raised to the power of infinity tends to 0" This is where you go wrong. Consider that, for example, $$limleft(1-frac4nright)^n=e^{-4}$$ but $$limleft(1-frac4{sqrt n}right)^n=0$$ and $$limleft(1-frac4{n^2}right)^n=1$$ although $$limleft(1-frac4nright)=limleft(1-frac4{sqrt n}right)=limleft(1-frac4{n^2}right)=1$$
$endgroup$
– Did
Jan 22 at 17:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You need to work out what happens to
begin{align}
ln(2+sqrt{9+x})
&=ln(2+3sqrt{1+(x/9)},)\
&=ln(2+3(1+x/18+o(x)))\
&=ln(5+x/6+o(x))\
&=ln 5+ln(1+x/30+o(x))\
&=ln 5+x/30+o(x)
end{align}
Thus the logarithm of your function is
$$
frac{ln 5-ln(2+sqrt{9+x})}{sin x}=
frac{ln 5-ln 5-x/30+o(x)}{x+o(x)}
$$
and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.
Your method #2 is flawed: for instance
$$
lim_{xto0^+}(1-x)^{1/x}=e^{-1}
$$
and not $0$ as you claim.
answered Jan 22 at 17:50
egregegreg
184k1486205
$endgroup$
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
add a comment |
$begingroup$
The limit is similar to
$$lim_{tto0}(1+t)^{1/t}.$$
In the positives, you can rewrite
$$lim_{utoinfty}left(frac{u+1}uright)^u=e.$$
In the negatives,
$$lim_{uto-infty}left(frac{u+1}uright)^u=lim_{utoinfty}left(frac{u-1}uright)^{-u}=lim_{utoinfty}left(frac v{v+1}right)^{-v-1}
\=lim_{vto-infty}left(frac{v+1}vright)^vlim_{vtoinfty}left(frac v{v+1}right)^{-1}=e.$$
So your second reasoning doesn't hold.
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
$endgroup$
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to work out what happens to
begin{align}
ln(2+sqrt{9+x})
&=ln(2+3sqrt{1+(x/9)},)\
&=ln(2+3(1+x/18+o(x)))\
&=ln(5+x/6+o(x))\
&=ln 5+ln(1+x/30+o(x))\
&=ln 5+x/30+o(x)
end{align}
Thus the logarithm of your function is
$$
frac{ln 5-ln(2+sqrt{9+x})}{sin x}=
frac{ln 5-ln 5-x/30+o(x)}{x+o(x)}
$$
and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.
Your method #2 is flawed: for instance
$$
lim_{xto0^+}(1-x)^{1/x}=e^{-1}
$$
and not $0$ as you claim.
answered Jan 22 at 17:50
egregegreg
184k1486205
$endgroup$
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
add a comment |
$begingroup$
You need to work out what happens to
begin{align}
ln(2+sqrt{9+x})
&=ln(2+3sqrt{1+(x/9)},)\
&=ln(2+3(1+x/18+o(x)))\
&=ln(5+x/6+o(x))\
&=ln 5+ln(1+x/30+o(x))\
&=ln 5+x/30+o(x)
end{align}
Thus the logarithm of your function is
$$
frac{ln 5-ln(2+sqrt{9+x})}{sin x}=
frac{ln 5-ln 5-x/30+o(x)}{x+o(x)}
$$
and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.
Your method #2 is flawed: for instance
$$
lim_{xto0^+}(1-x)^{1/x}=e^{-1}
$$
and not $0$ as you claim.
answered Jan 22 at 17:50
egregegreg
184k1486205
$endgroup$
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
add a comment |
$begingroup$
You need to work out what happens to
begin{align}
ln(2+sqrt{9+x})
&=ln(2+3sqrt{1+(x/9)},)\
&=ln(2+3(1+x/18+o(x)))\
&=ln(5+x/6+o(x))\
&=ln 5+ln(1+x/30+o(x))\
&=ln 5+x/30+o(x)
end{align}
Thus the logarithm of your function is
$$
frac{ln 5-ln(2+sqrt{9+x})}{sin x}=
frac{ln 5-ln 5-x/30+o(x)}{x+o(x)}
$$
and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.
Your method #2 is flawed: for instance
$$
lim_{xto0^+}(1-x)^{1/x}=e^{-1}
$$
and not $0$ as you claim.
answered Jan 22 at 17:50
egregegreg
184k1486205
$endgroup$
You need to work out what happens to
begin{align}
ln(2+sqrt{9+x})
&=ln(2+3sqrt{1+(x/9)},)\
&=ln(2+3(1+x/18+o(x)))\
&=ln(5+x/6+o(x))\
&=ln 5+ln(1+x/30+o(x))\
&=ln 5+x/30+o(x)
end{align}
Thus the logarithm of your function is
$$
frac{ln 5-ln(2+sqrt{9+x})}{sin x}=
frac{ln 5-ln 5-x/30+o(x)}{x+o(x)}
$$
and the limit (of the logarithm) is $-1/30$. So the given limit is $e^{-1/30}$. Good job with method #1.
Your method #2 is flawed: for instance
$$
lim_{xto0^+}(1-x)^{1/x}=e^{-1}
$$
and not $0$ as you claim.
answered Jan 22 at 17:50
egregegreg
184k1486205
answered Jan 22 at 17:50
egregegreg
184k1486205
answered Jan 22 at 17:50
egregegreg
184k1486205
answered Jan 22 at 17:50
egregegreg
184k1486205
184k1486205
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
add a comment |
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
$begingroup$
Thanks for the explanation. But can you please elaborate more on why the second method is wrong. I mean that why is a number less than 1 raised to the power of infinity not equal to zero..
$endgroup$
– Pranay
Feb 13 at 16:37
add a comment |
$begingroup$
The limit is similar to
$$lim_{tto0}(1+t)^{1/t}.$$
In the positives, you can rewrite
$$lim_{utoinfty}left(frac{u+1}uright)^u=e.$$
In the negatives,
$$lim_{uto-infty}left(frac{u+1}uright)^u=lim_{utoinfty}left(frac{u-1}uright)^{-u}=lim_{utoinfty}left(frac v{v+1}right)^{-v-1}
\=lim_{vto-infty}left(frac{v+1}vright)^vlim_{vtoinfty}left(frac v{v+1}right)^{-1}=e.$$
So your second reasoning doesn't hold.
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
$endgroup$
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
add a comment |
$begingroup$
The limit is similar to
$$lim_{tto0}(1+t)^{1/t}.$$
In the positives, you can rewrite
$$lim_{utoinfty}left(frac{u+1}uright)^u=e.$$
In the negatives,
$$lim_{uto-infty}left(frac{u+1}uright)^u=lim_{utoinfty}left(frac{u-1}uright)^{-u}=lim_{utoinfty}left(frac v{v+1}right)^{-v-1}
\=lim_{vto-infty}left(frac{v+1}vright)^vlim_{vtoinfty}left(frac v{v+1}right)^{-1}=e.$$
So your second reasoning doesn't hold.
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
$endgroup$
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
add a comment |
$begingroup$
The limit is similar to
$$lim_{tto0}(1+t)^{1/t}.$$
In the positives, you can rewrite
$$lim_{utoinfty}left(frac{u+1}uright)^u=e.$$
In the negatives,
$$lim_{uto-infty}left(frac{u+1}uright)^u=lim_{utoinfty}left(frac{u-1}uright)^{-u}=lim_{utoinfty}left(frac v{v+1}right)^{-v-1}
\=lim_{vto-infty}left(frac{v+1}vright)^vlim_{vtoinfty}left(frac v{v+1}right)^{-1}=e.$$
So your second reasoning doesn't hold.
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
$endgroup$
The limit is similar to
$$lim_{tto0}(1+t)^{1/t}.$$
In the positives, you can rewrite
$$lim_{utoinfty}left(frac{u+1}uright)^u=e.$$
In the negatives,
$$lim_{uto-infty}left(frac{u+1}uright)^u=lim_{utoinfty}left(frac{u-1}uright)^{-u}=lim_{utoinfty}left(frac v{v+1}right)^{-v-1}
\=lim_{vto-infty}left(frac{v+1}vright)^vlim_{vtoinfty}left(frac v{v+1}right)^{-1}=e.$$
So your second reasoning doesn't hold.
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
answered Jan 22 at 17:12
Yves DaoustYves Daoust
130k676227
130k676227
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
add a comment |
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
$begingroup$
Stupid downvote.
$endgroup$
– Yves Daoust
Jan 22 at 21:35
add a comment |
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$begingroup$
Welcome to Math StackExchange. While your question is good, please change your question title to describe your doubt better.
$endgroup$
– Swapnil Rustagi
Jan 22 at 16:54
$begingroup$
Please consider using left( and right) to get brackets that match the height of what's between them: $$left(frac{5}{x^2}right)$$ instead of $$(frac{5}{x^2})$$
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– Fly by Night
Jan 22 at 17:17
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"a number less than 1 raised to the power of infinity tends to 0" This is where you go wrong. Consider that, for example, $$limleft(1-frac4nright)^n=e^{-4}$$ but $$limleft(1-frac4{sqrt n}right)^n=0$$ and $$limleft(1-frac4{n^2}right)^n=1$$ although $$limleft(1-frac4nright)=limleft(1-frac4{sqrt n}right)=limleft(1-frac4{n^2}right)=1$$
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– Did
Jan 22 at 17:20