Mixing
Evaluate $limlimits_{xto infty} frac {x^x-4}{2^x-x^2}$
$begingroup$
Evaluate $$lim_{xto infty} frac {x^x-4}{2^x-x^2}$$
I think it needs to use L'Hospital Rule.
So, I first calculate $frac {d x^x}{dx}=
x^x(ln x+1)$.
And then $$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(ln x+1)}{2^x(ln 2)-2x}$$
It seems that I need to use L'Hospital Rule again. But when I do it, the thing inside the limit becomes more complicated.
How should I do next? Or maybe my way is false?
limits
edited Jan 9 at 13:14
user376343
3,4383827
asked Jan 9 at 12:52
MaggieMaggie
938
$endgroup$
add a comment |
$begingroup$
Evaluate $$lim_{xto infty} frac {x^x-4}{2^x-x^2}$$
I think it needs to use L'Hospital Rule.
So, I first calculate $frac {d x^x}{dx}=
x^x(ln x+1)$.
And then $$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(ln x+1)}{2^x(ln 2)-2x}$$
It seems that I need to use L'Hospital Rule again. But when I do it, the thing inside the limit becomes more complicated.
How should I do next? Or maybe my way is false?
limits
edited Jan 9 at 13:14
user376343
3,4383827
asked Jan 9 at 12:52
MaggieMaggie
938
$endgroup$
add a comment |
$begingroup$
Evaluate $$lim_{xto infty} frac {x^x-4}{2^x-x^2}$$
I think it needs to use L'Hospital Rule.
So, I first calculate $frac {d x^x}{dx}=
x^x(ln x+1)$.
And then $$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(ln x+1)}{2^x(ln 2)-2x}$$
It seems that I need to use L'Hospital Rule again. But when I do it, the thing inside the limit becomes more complicated.
How should I do next? Or maybe my way is false?
limits
edited Jan 9 at 13:14
user376343
3,4383827
asked Jan 9 at 12:52
MaggieMaggie
938
$endgroup$
Evaluate $$lim_{xto infty} frac {x^x-4}{2^x-x^2}$$
I think it needs to use L'Hospital Rule.
So, I first calculate $frac {d x^x}{dx}=
x^x(ln x+1)$.
And then $$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(ln x+1)}{2^x(ln 2)-2x}$$
It seems that I need to use L'Hospital Rule again. But when I do it, the thing inside the limit becomes more complicated.
How should I do next? Or maybe my way is false?
limits
limits
edited Jan 9 at 13:14
user376343
3,4383827
asked Jan 9 at 12:52
MaggieMaggie
938
edited Jan 9 at 13:14
user376343
3,4383827
asked Jan 9 at 12:52
MaggieMaggie
938
edited Jan 9 at 13:14
user376343
3,4383827
edited Jan 9 at 13:14
user376343
3,4383827
edited Jan 9 at 13:14
user376343
3,4383827
3,4383827
asked Jan 9 at 12:52
MaggieMaggie
938
asked Jan 9 at 12:52
MaggieMaggie
938
asked Jan 9 at 12:52
MaggieMaggie
938
938
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
$$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(1-frac4{x^x})}{2^x(1-frac{x^2}{2^x})}=lim_{xto infty}Big(frac x2Big)^xcdotBigg[frac{1-frac4{x^x}}{1-frac{x^2}{2^x}}Bigg]toinfty$$The latter tends to $1$, while the former tends to $infty$.
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
$endgroup$
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
2
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
add a comment |
$begingroup$
We can say that $4ll x^x$ and $x^2ll2^x$ as $x$ grows large so
$$lim_{xto infty} frac {x^x-4}{2^x-x^2} sim lim_{xto infty}frac{x^x}{2^x} = lim_{xto infty} (frac{x}{2})^x = infty$$
edited Jan 9 at 13:01
John Doe
11.1k11238
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
$endgroup$
add a comment |
$begingroup$
You’re better off avoiding L’Hôpital when solving limits. Keep it simple and divide all the terms by $x^x$:
$$lim_{xto infty} frac {1-frac{4}{x^x}}{frac{2^x}{x^x}-frac{x^2}{x^x}}$$
And now, you can tell that $frac{4}{x^x}$, $frac{2^x}{x^x}$, and $frac{x^2}{x^x}$ all tend to $0$, and since $2^x > x^2$ as $x$ grows large, the denominator tends to $0^+$. Hence, the limit becomes $frac{1}{0^+} = +infty$.
answered Jan 9 at 13:00
KM101KM101
5,9481524
$endgroup$
add a comment |
$begingroup$
Quick & dirty (though correct) method:
$4$ and $x^2$ are neglectible in front of the exponentials. Then the expression is asymptotic to
$$left(frac x2right)^x$$ which grows unboundedly.
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
Hint
for $x$ being sufficiently large ($x>5$) we have
$$2^x-x^2<2^x$$and $$x^x-4>{x^xover 2}$$therefore$${x^x-4over 2^x-x^2}>{1over 2}left({xover 2}right)^xto infty$$
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
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oldest
votes
$begingroup$
$$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(1-frac4{x^x})}{2^x(1-frac{x^2}{2^x})}=lim_{xto infty}Big(frac x2Big)^xcdotBigg[frac{1-frac4{x^x}}{1-frac{x^2}{2^x}}Bigg]toinfty$$The latter tends to $1$, while the former tends to $infty$.
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
$endgroup$
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
2
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
add a comment |
$begingroup$
$$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(1-frac4{x^x})}{2^x(1-frac{x^2}{2^x})}=lim_{xto infty}Big(frac x2Big)^xcdotBigg[frac{1-frac4{x^x}}{1-frac{x^2}{2^x}}Bigg]toinfty$$The latter tends to $1$, while the former tends to $infty$.
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
$endgroup$
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
2
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
add a comment |
$begingroup$
$$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(1-frac4{x^x})}{2^x(1-frac{x^2}{2^x})}=lim_{xto infty}Big(frac x2Big)^xcdotBigg[frac{1-frac4{x^x}}{1-frac{x^2}{2^x}}Bigg]toinfty$$The latter tends to $1$, while the former tends to $infty$.
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
$endgroup$
$$lim_{xto infty} frac {x^x-4}{2^x-x^2}=lim_{xto infty} frac {x^x(1-frac4{x^x})}{2^x(1-frac{x^2}{2^x})}=lim_{xto infty}Big(frac x2Big)^xcdotBigg[frac{1-frac4{x^x}}{1-frac{x^2}{2^x}}Bigg]toinfty$$The latter tends to $1$, while the former tends to $infty$.
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
edited Jan 9 at 13:06
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 12:57
Shubham JohriShubham Johri
5,122717
5,122717
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
2
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
add a comment |
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
2
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
$begingroup$
How to evaluate $lim_{xto infty} frac {1-frac {4}{x^x}}{1-frac {x^2}{2^x}}$
$endgroup$
– Maggie
Jan 9 at 13:06
2
2
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
$begingroup$
Both $frac{4}{x^x}$ and $frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $frac{1}{1} = 1$.
$endgroup$
– KM101
Jan 9 at 13:07
add a comment |
$begingroup$
We can say that $4ll x^x$ and $x^2ll2^x$ as $x$ grows large so
$$lim_{xto infty} frac {x^x-4}{2^x-x^2} sim lim_{xto infty}frac{x^x}{2^x} = lim_{xto infty} (frac{x}{2})^x = infty$$
edited Jan 9 at 13:01
John Doe
11.1k11238
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
$endgroup$
add a comment |
$begingroup$
We can say that $4ll x^x$ and $x^2ll2^x$ as $x$ grows large so
$$lim_{xto infty} frac {x^x-4}{2^x-x^2} sim lim_{xto infty}frac{x^x}{2^x} = lim_{xto infty} (frac{x}{2})^x = infty$$
edited Jan 9 at 13:01
John Doe
11.1k11238
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
$endgroup$
add a comment |
$begingroup$
We can say that $4ll x^x$ and $x^2ll2^x$ as $x$ grows large so
$$lim_{xto infty} frac {x^x-4}{2^x-x^2} sim lim_{xto infty}frac{x^x}{2^x} = lim_{xto infty} (frac{x}{2})^x = infty$$
edited Jan 9 at 13:01
John Doe
11.1k11238
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
$endgroup$
We can say that $4ll x^x$ and $x^2ll2^x$ as $x$ grows large so
$$lim_{xto infty} frac {x^x-4}{2^x-x^2} sim lim_{xto infty}frac{x^x}{2^x} = lim_{xto infty} (frac{x}{2})^x = infty$$
edited Jan 9 at 13:01
John Doe
11.1k11238
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
edited Jan 9 at 13:01
John Doe
11.1k11238
edited Jan 9 at 13:01
John Doe
11.1k11238
edited Jan 9 at 13:01
John Doe
11.1k11238
11.1k11238
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
answered Jan 9 at 12:59
Ahmad BazziAhmad Bazzi
8,0362724
8,0362724
add a comment |
add a comment |
$begingroup$
You’re better off avoiding L’Hôpital when solving limits. Keep it simple and divide all the terms by $x^x$:
$$lim_{xto infty} frac {1-frac{4}{x^x}}{frac{2^x}{x^x}-frac{x^2}{x^x}}$$
And now, you can tell that $frac{4}{x^x}$, $frac{2^x}{x^x}$, and $frac{x^2}{x^x}$ all tend to $0$, and since $2^x > x^2$ as $x$ grows large, the denominator tends to $0^+$. Hence, the limit becomes $frac{1}{0^+} = +infty$.
answered Jan 9 at 13:00
KM101KM101
5,9481524
$endgroup$
add a comment |
$begingroup$
You’re better off avoiding L’Hôpital when solving limits. Keep it simple and divide all the terms by $x^x$:
$$lim_{xto infty} frac {1-frac{4}{x^x}}{frac{2^x}{x^x}-frac{x^2}{x^x}}$$
And now, you can tell that $frac{4}{x^x}$, $frac{2^x}{x^x}$, and $frac{x^2}{x^x}$ all tend to $0$, and since $2^x > x^2$ as $x$ grows large, the denominator tends to $0^+$. Hence, the limit becomes $frac{1}{0^+} = +infty$.
answered Jan 9 at 13:00
KM101KM101
5,9481524
$endgroup$
add a comment |
$begingroup$
You’re better off avoiding L’Hôpital when solving limits. Keep it simple and divide all the terms by $x^x$:
$$lim_{xto infty} frac {1-frac{4}{x^x}}{frac{2^x}{x^x}-frac{x^2}{x^x}}$$
And now, you can tell that $frac{4}{x^x}$, $frac{2^x}{x^x}$, and $frac{x^2}{x^x}$ all tend to $0$, and since $2^x > x^2$ as $x$ grows large, the denominator tends to $0^+$. Hence, the limit becomes $frac{1}{0^+} = +infty$.
answered Jan 9 at 13:00
KM101KM101
5,9481524
$endgroup$
You’re better off avoiding L’Hôpital when solving limits. Keep it simple and divide all the terms by $x^x$:
$$lim_{xto infty} frac {1-frac{4}{x^x}}{frac{2^x}{x^x}-frac{x^2}{x^x}}$$
And now, you can tell that $frac{4}{x^x}$, $frac{2^x}{x^x}$, and $frac{x^2}{x^x}$ all tend to $0$, and since $2^x > x^2$ as $x$ grows large, the denominator tends to $0^+$. Hence, the limit becomes $frac{1}{0^+} = +infty$.
answered Jan 9 at 13:00
KM101KM101
5,9481524
answered Jan 9 at 13:00
KM101KM101
5,9481524
answered Jan 9 at 13:00
KM101KM101
5,9481524
answered Jan 9 at 13:00
KM101KM101
5,9481524
5,9481524
add a comment |
add a comment |
$begingroup$
Quick & dirty (though correct) method:
$4$ and $x^2$ are neglectible in front of the exponentials. Then the expression is asymptotic to
$$left(frac x2right)^x$$ which grows unboundedly.
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
Quick & dirty (though correct) method:
$4$ and $x^2$ are neglectible in front of the exponentials. Then the expression is asymptotic to
$$left(frac x2right)^x$$ which grows unboundedly.
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
$begingroup$
Quick & dirty (though correct) method:
$4$ and $x^2$ are neglectible in front of the exponentials. Then the expression is asymptotic to
$$left(frac x2right)^x$$ which grows unboundedly.
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
$endgroup$
Quick & dirty (though correct) method:
$4$ and $x^2$ are neglectible in front of the exponentials. Then the expression is asymptotic to
$$left(frac x2right)^x$$ which grows unboundedly.
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
edited Jan 9 at 13:07
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
answered Jan 9 at 12:59
Yves DaoustYves Daoust
126k672226
126k672226
add a comment |
add a comment |
$begingroup$
Hint
for $x$ being sufficiently large ($x>5$) we have
$$2^x-x^2<2^x$$and $$x^x-4>{x^xover 2}$$therefore$${x^x-4over 2^x-x^2}>{1over 2}left({xover 2}right)^xto infty$$
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
Hint
for $x$ being sufficiently large ($x>5$) we have
$$2^x-x^2<2^x$$and $$x^x-4>{x^xover 2}$$therefore$${x^x-4over 2^x-x^2}>{1over 2}left({xover 2}right)^xto infty$$
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
Hint
for $x$ being sufficiently large ($x>5$) we have
$$2^x-x^2<2^x$$and $$x^x-4>{x^xover 2}$$therefore$${x^x-4over 2^x-x^2}>{1over 2}left({xover 2}right)^xto infty$$
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
Hint
for $x$ being sufficiently large ($x>5$) we have
$$2^x-x^2<2^x$$and $$x^x-4>{x^xover 2}$$therefore$${x^x-4over 2^x-x^2}>{1over 2}left({xover 2}right)^xto infty$$
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 16:05
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
add a comment |
add a comment |
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