What is the error in this approach to evaluating the limit? $lim_xto0(sinsin x- xsqrt[3]1-x^2)/x^5$Analyzing limits problem Calculus (tell me where I'm wrong).Extended limit laws for step by step evaluation of limitsEvaluating $lim_xto0(xtan x)^x$Evaluating $displaystyle lim_xto0 fractan(2x)sin x$Find $limlimits_xto0fracsqrt1+tan x-sqrt1+xsin^2x$Find $lim_xto0fracsinleft(1-fracsin(x)xright)x^2$. Is my approach correct?Calculate the Limit as x approaches 0: $lim_xto0fracln(1+sin x)sin(2x)$Find $lim_xto0fracln(1+sin^3x cos^2x)cot(ln^3(1+x))tan^4xsin(sqrtx^2+2-sqrt2)ln(1+x^2)$Limit of $lim_xto0^+fracsin xsin sqrtx$Evaluating the limit $lim_limitsxto0^+frace^x-cos(lambda sqrt x)sqrt 1+sin(lambda x)-1$Limit $lim_limitsxto0tanx-sinxover x^3$Limit $lim_limitsxto02x-sinxover3x+sinx$
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What is the error in this approach to evaluating the limit? $lim_xto0(sinsin x- xsqrt[3]1-x^2)/x^5$
Analyzing limits problem Calculus (tell me where I'm wrong).Extended limit laws for step by step evaluation of limitsEvaluating $lim_xto0(xtan x)^x$Evaluating $displaystyle lim_xto0 fractan(2x)sin x$Find $limlimits_xto0fracsqrt1+tan x-sqrt1+xsin^2x$Find $lim_xto0fracsinleft(1-fracsin(x)xright)x^2$. Is my approach correct?Calculate the Limit as x approaches 0: $lim_xto0fracln(1+sin x)sin(2x)$Find $lim_xto0fracln(1+sin^3x cos^2x)cot(ln^3(1+x))tan^4xsin(sqrtx^2+2-sqrt2)ln(1+x^2)$Limit of $lim_xto0^+fracsin xsin sqrtx$Evaluating the limit $lim_limitsxto0^+frace^x-cos(lambda sqrt x)sqrt 1+sin(lambda x)-1$Limit $lim_limitsxto0tanx-sinxover x^3$Limit $lim_limitsxto02x-sinxover3x+sinx$
$begingroup$
I am having trouble with a small thing in a limit. I encountered the limit
$$lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5$$
which can be solved rather easily by using Taylor expansion and calculating the quotient. The problem, though, arised when I tried the more natural approach - I tried to simplify the expression. When doing that, I simplified the fraction by separating it into the two terms, first of which is $fracsinsin xx^5$. Then, I multiplied this term by $fracsin xsin x$ which creates $fracsinsin xsin x×fracsin xx^5$. When trying to find the original limit with this expression, it is still the same as the original limit should be. Although, I noticed that when I continue by firstly taking $lim_xto0fracsinsin xsin x=1$, the overall limit then simplifies to
$$lim_xto0fracsin xx^5-fracsqrt[3]1-x^2x^4$$
This limit, though, is equal to $infty$, which is contrary to the original result, which should be $frac1990$ (by using Taylor). Can anyone tell me what creates this error? I was thinking maybe the multiplying of $fracsin xsin x$, but since that is just $1$ (no limits taken there), it should not pose a problem. The problem just seems to arise when I use the rule $lim f(x)g(x)= lim f(x)lim g(x)$ on the first term there. Is this rule invalid in this case? If so, why? What am I missing? Sorry if this is really trivial, I am pretty tired by now and am probably missing something stupidly simple, but I really don't see it.
In advance, thanks for your time.
limits trigonometry
edited Mar 13 at 17:43
6005
37k751127
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am having trouble with a small thing in a limit. I encountered the limit
$$lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5$$
which can be solved rather easily by using Taylor expansion and calculating the quotient. The problem, though, arised when I tried the more natural approach - I tried to simplify the expression. When doing that, I simplified the fraction by separating it into the two terms, first of which is $fracsinsin xx^5$. Then, I multiplied this term by $fracsin xsin x$ which creates $fracsinsin xsin x×fracsin xx^5$. When trying to find the original limit with this expression, it is still the same as the original limit should be. Although, I noticed that when I continue by firstly taking $lim_xto0fracsinsin xsin x=1$, the overall limit then simplifies to
$$lim_xto0fracsin xx^5-fracsqrt[3]1-x^2x^4$$
This limit, though, is equal to $infty$, which is contrary to the original result, which should be $frac1990$ (by using Taylor). Can anyone tell me what creates this error? I was thinking maybe the multiplying of $fracsin xsin x$, but since that is just $1$ (no limits taken there), it should not pose a problem. The problem just seems to arise when I use the rule $lim f(x)g(x)= lim f(x)lim g(x)$ on the first term there. Is this rule invalid in this case? If so, why? What am I missing? Sorry if this is really trivial, I am pretty tired by now and am probably missing something stupidly simple, but I really don't see it.
In advance, thanks for your time.
limits trigonometry
edited Mar 13 at 17:43
6005
37k751127
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
This is a typical mistake. A sub-expression can be replaced by its limit only if it occurs as a factor or a term in the original expression.
$endgroup$
– Paramanand Singh
Mar 13 at 17:37
1
$begingroup$
See more details at math.stackexchange.com/a/1783818/72031 and math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Mar 13 at 17:39
$begingroup$
thank you, I understand now.
$endgroup$
– xDeeZee
Mar 13 at 18:32
add a comment |
$begingroup$
I am having trouble with a small thing in a limit. I encountered the limit
$$lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5$$
which can be solved rather easily by using Taylor expansion and calculating the quotient. The problem, though, arised when I tried the more natural approach - I tried to simplify the expression. When doing that, I simplified the fraction by separating it into the two terms, first of which is $fracsinsin xx^5$. Then, I multiplied this term by $fracsin xsin x$ which creates $fracsinsin xsin x×fracsin xx^5$. When trying to find the original limit with this expression, it is still the same as the original limit should be. Although, I noticed that when I continue by firstly taking $lim_xto0fracsinsin xsin x=1$, the overall limit then simplifies to
$$lim_xto0fracsin xx^5-fracsqrt[3]1-x^2x^4$$
This limit, though, is equal to $infty$, which is contrary to the original result, which should be $frac1990$ (by using Taylor). Can anyone tell me what creates this error? I was thinking maybe the multiplying of $fracsin xsin x$, but since that is just $1$ (no limits taken there), it should not pose a problem. The problem just seems to arise when I use the rule $lim f(x)g(x)= lim f(x)lim g(x)$ on the first term there. Is this rule invalid in this case? If so, why? What am I missing? Sorry if this is really trivial, I am pretty tired by now and am probably missing something stupidly simple, but I really don't see it.
In advance, thanks for your time.
limits trigonometry
edited Mar 13 at 17:43
6005
37k751127
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am having trouble with a small thing in a limit. I encountered the limit
$$lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5$$
which can be solved rather easily by using Taylor expansion and calculating the quotient. The problem, though, arised when I tried the more natural approach - I tried to simplify the expression. When doing that, I simplified the fraction by separating it into the two terms, first of which is $fracsinsin xx^5$. Then, I multiplied this term by $fracsin xsin x$ which creates $fracsinsin xsin x×fracsin xx^5$. When trying to find the original limit with this expression, it is still the same as the original limit should be. Although, I noticed that when I continue by firstly taking $lim_xto0fracsinsin xsin x=1$, the overall limit then simplifies to
$$lim_xto0fracsin xx^5-fracsqrt[3]1-x^2x^4$$
This limit, though, is equal to $infty$, which is contrary to the original result, which should be $frac1990$ (by using Taylor). Can anyone tell me what creates this error? I was thinking maybe the multiplying of $fracsin xsin x$, but since that is just $1$ (no limits taken there), it should not pose a problem. The problem just seems to arise when I use the rule $lim f(x)g(x)= lim f(x)lim g(x)$ on the first term there. Is this rule invalid in this case? If so, why? What am I missing? Sorry if this is really trivial, I am pretty tired by now and am probably missing something stupidly simple, but I really don't see it.
In advance, thanks for your time.
limits trigonometry
limits trigonometry
edited Mar 13 at 17:43
6005
37k751127
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 13 at 17:43
6005
37k751127
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 13 at 17:43
6005
37k751127
edited Mar 13 at 17:43
6005
37k751127
edited Mar 13 at 17:43
6005
37k751127
37k751127
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
asked Mar 13 at 17:24
xDeeZeexDeeZee
62
62
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
xDeeZee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
This is a typical mistake. A sub-expression can be replaced by its limit only if it occurs as a factor or a term in the original expression.
$endgroup$
– Paramanand Singh
Mar 13 at 17:37
1
$begingroup$
See more details at math.stackexchange.com/a/1783818/72031 and math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Mar 13 at 17:39
$begingroup$
thank you, I understand now.
$endgroup$
– xDeeZee
Mar 13 at 18:32
add a comment |
2
$begingroup$
This is a typical mistake. A sub-expression can be replaced by its limit only if it occurs as a factor or a term in the original expression.
$endgroup$
– Paramanand Singh
Mar 13 at 17:37
1
$begingroup$
See more details at math.stackexchange.com/a/1783818/72031 and math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Mar 13 at 17:39
$begingroup$
thank you, I understand now.
$endgroup$
– xDeeZee
Mar 13 at 18:32
2
2
$begingroup$
This is a typical mistake. A sub-expression can be replaced by its limit only if it occurs as a factor or a term in the original expression.
$endgroup$
– Paramanand Singh
Mar 13 at 17:37
$begingroup$
This is a typical mistake. A sub-expression can be replaced by its limit only if it occurs as a factor or a term in the original expression.
$endgroup$
– Paramanand Singh
Mar 13 at 17:37
1
1
$begingroup$
See more details at math.stackexchange.com/a/1783818/72031 and math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Mar 13 at 17:39
$begingroup$
See more details at math.stackexchange.com/a/1783818/72031 and math.stackexchange.com/q/2971122/72031
$endgroup$
– Paramanand Singh
Mar 13 at 17:39
$begingroup$
thank you, I understand now.
$endgroup$
– xDeeZee
Mar 13 at 18:32
$begingroup$
thank you, I understand now.
$endgroup$
– xDeeZee
Mar 13 at 18:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The problem is that you implicitly also applied the sum-of-limits rule:
$$
lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5 = lim_x to 0 left( fracsinsin xx^5 right) - lim_x to 0 left( fracx sqrt[3]1 - x^2x^5 right).
$$
However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $infty - infty$ is an indeterminate form.)
It's necessary to apply this rule first before doing your trick of multiplying by $fracsin xsin x$. Otherwise, you can't "distribute" the $fracsin xsin x$ into the first term only -- you would have to multiply the second term by $fracsin xsin x$ as well.
For a simpler example of what went wrong, consider
$$
lim_x to infty (x - x) = 0.
$$
We can write this as
$$
lim_x to infty left( fracx+1x+1 cdot x - xright)
= lim_x to infty left( fracxx+1 cdot (x+1) - xright),
$$
then according to the logic you used, since $fracxx+1 to 1$, we can simplify this to
$$
lim_x to infty left( (x + 1) - xright) = 1.
$$
Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.
answered Mar 13 at 17:39
60056005
37k751127
$endgroup$
$begingroup$
thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
$endgroup$
– xDeeZee
Mar 13 at 18:25
$begingroup$
@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
$endgroup$
– 6005
Mar 13 at 18:34
$begingroup$
All good, then. I understand the problem.
$endgroup$
– xDeeZee
Mar 13 at 18:39
add a comment |
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$begingroup$
The problem is that you implicitly also applied the sum-of-limits rule:
$$
lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5 = lim_x to 0 left( fracsinsin xx^5 right) - lim_x to 0 left( fracx sqrt[3]1 - x^2x^5 right).
$$
However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $infty - infty$ is an indeterminate form.)
It's necessary to apply this rule first before doing your trick of multiplying by $fracsin xsin x$. Otherwise, you can't "distribute" the $fracsin xsin x$ into the first term only -- you would have to multiply the second term by $fracsin xsin x$ as well.
For a simpler example of what went wrong, consider
$$
lim_x to infty (x - x) = 0.
$$
We can write this as
$$
lim_x to infty left( fracx+1x+1 cdot x - xright)
= lim_x to infty left( fracxx+1 cdot (x+1) - xright),
$$
then according to the logic you used, since $fracxx+1 to 1$, we can simplify this to
$$
lim_x to infty left( (x + 1) - xright) = 1.
$$
Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.
answered Mar 13 at 17:39
60056005
37k751127
$endgroup$
$begingroup$
thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
$endgroup$
– xDeeZee
Mar 13 at 18:25
$begingroup$
@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
$endgroup$
– 6005
Mar 13 at 18:34
$begingroup$
All good, then. I understand the problem.
$endgroup$
– xDeeZee
Mar 13 at 18:39
add a comment |
$begingroup$
The problem is that you implicitly also applied the sum-of-limits rule:
$$
lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5 = lim_x to 0 left( fracsinsin xx^5 right) - lim_x to 0 left( fracx sqrt[3]1 - x^2x^5 right).
$$
However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $infty - infty$ is an indeterminate form.)
It's necessary to apply this rule first before doing your trick of multiplying by $fracsin xsin x$. Otherwise, you can't "distribute" the $fracsin xsin x$ into the first term only -- you would have to multiply the second term by $fracsin xsin x$ as well.
For a simpler example of what went wrong, consider
$$
lim_x to infty (x - x) = 0.
$$
We can write this as
$$
lim_x to infty left( fracx+1x+1 cdot x - xright)
= lim_x to infty left( fracxx+1 cdot (x+1) - xright),
$$
then according to the logic you used, since $fracxx+1 to 1$, we can simplify this to
$$
lim_x to infty left( (x + 1) - xright) = 1.
$$
Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.
answered Mar 13 at 17:39
60056005
37k751127
$endgroup$
$begingroup$
thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
$endgroup$
– xDeeZee
Mar 13 at 18:25
$begingroup$
@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
$endgroup$
– 6005
Mar 13 at 18:34
$begingroup$
All good, then. I understand the problem.
$endgroup$
– xDeeZee
Mar 13 at 18:39
add a comment |
$begingroup$
The problem is that you implicitly also applied the sum-of-limits rule:
$$
lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5 = lim_x to 0 left( fracsinsin xx^5 right) - lim_x to 0 left( fracx sqrt[3]1 - x^2x^5 right).
$$
However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $infty - infty$ is an indeterminate form.)
It's necessary to apply this rule first before doing your trick of multiplying by $fracsin xsin x$. Otherwise, you can't "distribute" the $fracsin xsin x$ into the first term only -- you would have to multiply the second term by $fracsin xsin x$ as well.
For a simpler example of what went wrong, consider
$$
lim_x to infty (x - x) = 0.
$$
We can write this as
$$
lim_x to infty left( fracx+1x+1 cdot x - xright)
= lim_x to infty left( fracxx+1 cdot (x+1) - xright),
$$
then according to the logic you used, since $fracxx+1 to 1$, we can simplify this to
$$
lim_x to infty left( (x + 1) - xright) = 1.
$$
Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.
answered Mar 13 at 17:39
60056005
37k751127
$endgroup$
The problem is that you implicitly also applied the sum-of-limits rule:
$$
lim_xto0fracsinsin x- xsqrt[3]1-x^2x^5 = lim_x to 0 left( fracsinsin xx^5 right) - lim_x to 0 left( fracx sqrt[3]1 - x^2x^5 right).
$$
However, this rule does not work because the two limits do not exist as real numbers. (Or if you prefer: $infty - infty$ is an indeterminate form.)
It's necessary to apply this rule first before doing your trick of multiplying by $fracsin xsin x$. Otherwise, you can't "distribute" the $fracsin xsin x$ into the first term only -- you would have to multiply the second term by $fracsin xsin x$ as well.
For a simpler example of what went wrong, consider
$$
lim_x to infty (x - x) = 0.
$$
We can write this as
$$
lim_x to infty left( fracx+1x+1 cdot x - xright)
= lim_x to infty left( fracxx+1 cdot (x+1) - xright),
$$
then according to the logic you used, since $fracxx+1 to 1$, we can simplify this to
$$
lim_x to infty left( (x + 1) - xright) = 1.
$$
Essentially, the product-of-limit rules does not apply if you are only applying it to some subterm of the limit. For a limit of the form $lim (A + B)$, you can't apply the product-of-limit rule to just $A$ on its own.
answered Mar 13 at 17:39
60056005
37k751127
edited Mar 13 at 18:34
answered Mar 13 at 17:39
60056005
37k751127
answered Mar 13 at 17:39
60056005
37k751127
answered Mar 13 at 17:39
60056005
37k751127
37k751127
$begingroup$
thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
$endgroup$
– xDeeZee
Mar 13 at 18:25
$begingroup$
@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
$endgroup$
– 6005
Mar 13 at 18:34
$begingroup$
All good, then. I understand the problem.
$endgroup$
– xDeeZee
Mar 13 at 18:39
add a comment |
$begingroup$
thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
$endgroup$
– xDeeZee
Mar 13 at 18:25
$begingroup$
@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
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– 6005
Mar 13 at 18:34
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All good, then. I understand the problem.
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– xDeeZee
Mar 13 at 18:39
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thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
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– xDeeZee
Mar 13 at 18:25
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thank you very much for your answer, although, you refer that the sum of limits rule does not apply because the limits do not exist - they do, don't they? They are both infinity, but indeed exist. Or am I missing something?
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– xDeeZee
Mar 13 at 18:25
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@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
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– 6005
Mar 13 at 18:34
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@xDeeZee No, you aren't missing something, I meant exist as a real number (not an extended real number). But I will edit to make it clearer.
$endgroup$
– 6005
Mar 13 at 18:34
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All good, then. I understand the problem.
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– xDeeZee
Mar 13 at 18:39
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All good, then. I understand the problem.
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– xDeeZee
Mar 13 at 18:39
add a comment |
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This is a typical mistake. A sub-expression can be replaced by its limit only if it occurs as a factor or a term in the original expression.
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– Paramanand Singh
Mar 13 at 17:37
1
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See more details at math.stackexchange.com/a/1783818/72031 and math.stackexchange.com/q/2971122/72031
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– Paramanand Singh
Mar 13 at 17:39
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thank you, I understand now.
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– xDeeZee
Mar 13 at 18:32