Comparing two infinitely large functions
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I have $ f(x)=frac{x*arctgx}{sqrt{4x+3}} $
And $ g(x)=sqrt x - {x}^{1/3} $
Limits of both functions are infinity, so these functions are infinitely large .
I need to write equivalent functions for both as $ C{x}^{n} $ and find out order of growth/smallness of functions and then compare f(x) and g(x).
I don't no how to write equivalent for f(x) and what are common ideas of tasks like this, in my lecture copybook I didn't find anything/
calculus limits functions
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$begingroup$
I have $ f(x)=frac{x*arctgx}{sqrt{4x+3}} $
And $ g(x)=sqrt x - {x}^{1/3} $
Limits of both functions are infinity, so these functions are infinitely large .
I need to write equivalent functions for both as $ C{x}^{n} $ and find out order of growth/smallness of functions and then compare f(x) and g(x).
I don't no how to write equivalent for f(x) and what are common ideas of tasks like this, in my lecture copybook I didn't find anything/
calculus limits functions
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add a comment |
$begingroup$
I have $ f(x)=frac{x*arctgx}{sqrt{4x+3}} $
And $ g(x)=sqrt x - {x}^{1/3} $
Limits of both functions are infinity, so these functions are infinitely large .
I need to write equivalent functions for both as $ C{x}^{n} $ and find out order of growth/smallness of functions and then compare f(x) and g(x).
I don't no how to write equivalent for f(x) and what are common ideas of tasks like this, in my lecture copybook I didn't find anything/
calculus limits functions
$endgroup$
I have $ f(x)=frac{x*arctgx}{sqrt{4x+3}} $
And $ g(x)=sqrt x - {x}^{1/3} $
Limits of both functions are infinity, so these functions are infinitely large .
I need to write equivalent functions for both as $ C{x}^{n} $ and find out order of growth/smallness of functions and then compare f(x) and g(x).
I don't no how to write equivalent for f(x) and what are common ideas of tasks like this, in my lecture copybook I didn't find anything/
calculus limits functions
calculus limits functions
edited Jan 16 at 9:10
MaoWao
3,318617
3,318617
asked Jan 16 at 8:48


TovarischTovarisch
297
297
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$f(x)=x^{1/2} frac {arctan (x)} {sqrt {4+frac 3 x}}$ and $g(x)=x^{1/2} (1-x^{frac 1 3 -frac 1 2})$. So both functions behave like $x^{1/2}$ when $x to infty$ and $lim frac {f(x)} {g(x)}=frac {pi} 4$ as $x to infty$.
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I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
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– Toffomat
Jan 16 at 9:30
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@Toffomat Right. Thanks for pointing out.
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– Kavi Rama Murthy
Jan 16 at 9:32
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1 Answer
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1 Answer
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$begingroup$
$f(x)=x^{1/2} frac {arctan (x)} {sqrt {4+frac 3 x}}$ and $g(x)=x^{1/2} (1-x^{frac 1 3 -frac 1 2})$. So both functions behave like $x^{1/2}$ when $x to infty$ and $lim frac {f(x)} {g(x)}=frac {pi} 4$ as $x to infty$.
$endgroup$
$begingroup$
I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
$endgroup$
– Toffomat
Jan 16 at 9:30
$begingroup$
@Toffomat Right. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 9:32
add a comment |
$begingroup$
$f(x)=x^{1/2} frac {arctan (x)} {sqrt {4+frac 3 x}}$ and $g(x)=x^{1/2} (1-x^{frac 1 3 -frac 1 2})$. So both functions behave like $x^{1/2}$ when $x to infty$ and $lim frac {f(x)} {g(x)}=frac {pi} 4$ as $x to infty$.
$endgroup$
$begingroup$
I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
$endgroup$
– Toffomat
Jan 16 at 9:30
$begingroup$
@Toffomat Right. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 9:32
add a comment |
$begingroup$
$f(x)=x^{1/2} frac {arctan (x)} {sqrt {4+frac 3 x}}$ and $g(x)=x^{1/2} (1-x^{frac 1 3 -frac 1 2})$. So both functions behave like $x^{1/2}$ when $x to infty$ and $lim frac {f(x)} {g(x)}=frac {pi} 4$ as $x to infty$.
$endgroup$
$f(x)=x^{1/2} frac {arctan (x)} {sqrt {4+frac 3 x}}$ and $g(x)=x^{1/2} (1-x^{frac 1 3 -frac 1 2})$. So both functions behave like $x^{1/2}$ when $x to infty$ and $lim frac {f(x)} {g(x)}=frac {pi} 4$ as $x to infty$.
edited Jan 16 at 9:31
answered Jan 16 at 8:53


Kavi Rama MurthyKavi Rama Murthy
61.7k42262
61.7k42262
$begingroup$
I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
$endgroup$
– Toffomat
Jan 16 at 9:30
$begingroup$
@Toffomat Right. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 9:32
add a comment |
$begingroup$
I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
$endgroup$
– Toffomat
Jan 16 at 9:30
$begingroup$
@Toffomat Right. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 9:32
$begingroup$
I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
$endgroup$
– Toffomat
Jan 16 at 9:30
$begingroup$
I guess you forgot a square root in the denominator of $f(x)$. Then the limit will be $pi/4$.
$endgroup$
– Toffomat
Jan 16 at 9:30
$begingroup$
@Toffomat Right. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 9:32
$begingroup$
@Toffomat Right. Thanks for pointing out.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 9:32
add a comment |
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