Question on limits (0/0 form)
$begingroup$
Posted pic of question because i don't know how to type this question (specially roots)
Please tell me how to solve such questions..
Initially solved question ( of for 0/0,∞/∞ etc) by differentiating the numerator and denominator and then solve
But here it is getting complicated by differentiating method.
calculus limits
asked Jan 10 at 11:05
AkashAkash
786
$endgroup$
add a comment |
$begingroup$
Posted pic of question because i don't know how to type this question (specially roots)
Please tell me how to solve such questions..
Initially solved question ( of for 0/0,∞/∞ etc) by differentiating the numerator and denominator and then solve
But here it is getting complicated by differentiating method.
calculus limits
asked Jan 10 at 11:05
AkashAkash
786
$endgroup$
2
$begingroup$
The photo is sidewise
$endgroup$
– William Elliot
Jan 10 at 11:32
$begingroup$
I know, what is the problem in it.
$endgroup$
– Akash
Jan 10 at 11:39
$begingroup$
I don't understand what is you question. You are given a solution below the exercise. What is going there? Well $n$ gets factored out and canceled, while then by continuity of root: $limsqrt{1+{a over n}} = sqrt{limleft({1 + {a over n}}right)} = sqrt{1 + lim{aover n}}$. What is the limit of ${aover n}$ as $n to infty$ given $a in Bbb R$ is some fixed number? ${aover n}$ gets infinitely small and hence the root is tending to $1$ as $n$ tends to infinity.
$endgroup$
– roman
Jan 10 at 11:54
$begingroup$
Also please use this reference on how to typeset math on this site. There are several examples of typesetting roots in there
$endgroup$
– roman
Jan 10 at 11:57
add a comment |
$begingroup$
Posted pic of question because i don't know how to type this question (specially roots)
Please tell me how to solve such questions..
Initially solved question ( of for 0/0,∞/∞ etc) by differentiating the numerator and denominator and then solve
But here it is getting complicated by differentiating method.
calculus limits
asked Jan 10 at 11:05
AkashAkash
786
$endgroup$
Posted pic of question because i don't know how to type this question (specially roots)
Please tell me how to solve such questions..
Initially solved question ( of for 0/0,∞/∞ etc) by differentiating the numerator and denominator and then solve
But here it is getting complicated by differentiating method.
calculus limits
calculus limits
asked Jan 10 at 11:05
AkashAkash
786
asked Jan 10 at 11:05
AkashAkash
786
edited Jan 10 at 11:11
Akash
asked Jan 10 at 11:05
AkashAkash
786
asked Jan 10 at 11:05
AkashAkash
786
asked Jan 10 at 11:05
AkashAkash
786
786
2
$begingroup$
The photo is sidewise
$endgroup$
– William Elliot
Jan 10 at 11:32
$begingroup$
I know, what is the problem in it.
$endgroup$
– Akash
Jan 10 at 11:39
$begingroup$
I don't understand what is you question. You are given a solution below the exercise. What is going there? Well $n$ gets factored out and canceled, while then by continuity of root: $limsqrt{1+{a over n}} = sqrt{limleft({1 + {a over n}}right)} = sqrt{1 + lim{aover n}}$. What is the limit of ${aover n}$ as $n to infty$ given $a in Bbb R$ is some fixed number? ${aover n}$ gets infinitely small and hence the root is tending to $1$ as $n$ tends to infinity.
$endgroup$
– roman
Jan 10 at 11:54
$begingroup$
Also please use this reference on how to typeset math on this site. There are several examples of typesetting roots in there
$endgroup$
– roman
Jan 10 at 11:57
add a comment |
2
$begingroup$
The photo is sidewise
$endgroup$
– William Elliot
Jan 10 at 11:32
$begingroup$
I know, what is the problem in it.
$endgroup$
– Akash
Jan 10 at 11:39
$begingroup$
I don't understand what is you question. You are given a solution below the exercise. What is going there? Well $n$ gets factored out and canceled, while then by continuity of root: $limsqrt{1+{a over n}} = sqrt{limleft({1 + {a over n}}right)} = sqrt{1 + lim{aover n}}$. What is the limit of ${aover n}$ as $n to infty$ given $a in Bbb R$ is some fixed number? ${aover n}$ gets infinitely small and hence the root is tending to $1$ as $n$ tends to infinity.
$endgroup$
– roman
Jan 10 at 11:54
$begingroup$
Also please use this reference on how to typeset math on this site. There are several examples of typesetting roots in there
$endgroup$
– roman
Jan 10 at 11:57
2
2
$begingroup$
The photo is sidewise
$endgroup$
– William Elliot
Jan 10 at 11:32
$begingroup$
The photo is sidewise
$endgroup$
– William Elliot
Jan 10 at 11:32
$begingroup$
I know, what is the problem in it.
$endgroup$
– Akash
Jan 10 at 11:39
$begingroup$
I know, what is the problem in it.
$endgroup$
– Akash
Jan 10 at 11:39
$begingroup$
I don't understand what is you question. You are given a solution below the exercise. What is going there? Well $n$ gets factored out and canceled, while then by continuity of root: $limsqrt{1+{a over n}} = sqrt{limleft({1 + {a over n}}right)} = sqrt{1 + lim{aover n}}$. What is the limit of ${aover n}$ as $n to infty$ given $a in Bbb R$ is some fixed number? ${aover n}$ gets infinitely small and hence the root is tending to $1$ as $n$ tends to infinity.
$endgroup$
– roman
Jan 10 at 11:54
$begingroup$
I don't understand what is you question. You are given a solution below the exercise. What is going there? Well $n$ gets factored out and canceled, while then by continuity of root: $limsqrt{1+{a over n}} = sqrt{limleft({1 + {a over n}}right)} = sqrt{1 + lim{aover n}}$. What is the limit of ${aover n}$ as $n to infty$ given $a in Bbb R$ is some fixed number? ${aover n}$ gets infinitely small and hence the root is tending to $1$ as $n$ tends to infinity.
$endgroup$
– roman
Jan 10 at 11:54
$begingroup$
Also please use this reference on how to typeset math on this site. There are several examples of typesetting roots in there
$endgroup$
– roman
Jan 10 at 11:57
$begingroup$
Also please use this reference on how to typeset math on this site. There are several examples of typesetting roots in there
$endgroup$
– roman
Jan 10 at 11:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When using $ntoinfty$, it is generally assumed that we're dealing with a sequence; it doesn't really matter here, because the functions $xmapstosqrt{x^2+4x}$ and $xmapstosqrt[3]{x^3-3x^2}$ are monotonic in a neighborhood of $infty$.
L'Hôpital's theorem which you seem to refer to is one of the tools, but certainly not the tool of choice in every case. Indeed, if you try it, you're left with something like
$$
frac{x+2}{sqrt{x^2+4x}}frac{sqrt[3]{(x^3-3x^2)^2}}{x^2-2x}
$$
which is much worse than what you started with.
Another nice example is
$$
lim_{xtoinfty}frac{x}{sqrt{x^2+1}}overset{text{(H)}}{=}
lim_{xtoinfty}frac{1}{;dfrac{x}{sqrt{x^2+1}}}=
lim_{xtoinfty}frac{sqrt{x^2+1}}{x}
$$
and this loops.
In this case, collecting the largest power of $n$ is a good method; you can also do the substitution $n=1/t$ at the outset, getting
$$
lim_{tto0^+}frac{sqrt{dfrac{1}{t^2}+dfrac{4}{t}}}{sqrt[3]{dfrac{1}{t^3}-dfrac{3}{t^2}}}=
lim_{tto0^+}frac{dfrac{sqrt{1+4t}}{|t|}}{dfrac{sqrt[3]{1-3t}}{t}}=
lim_{tto0^+}frac{sqrt{1+4t}}{sqrt[3]{1-3t}}
$$
which is elementary.
answered Jan 10 at 12:12
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
Well, I'd start by suggesting that you try to follow the solution printed there. We write $sqrt{n^2+text{stuff}}=nsqrt{1+text{small stuff}}$ and $sqrt[3]{n^3+text{stuff}}=nsqrt[3]{1+text{small stuff}}$. As $ntoinfty$, the parts I've labeled "small stuff" go to zero, and the limiting ratio is $frac{nsqrt{1}}{nsqrt[3]{1}}=frac nn=1$.
When dealing with limits, the first instinct should be to focus on what's big. $n^2$ is much bigger than $4n$ as $ntoinfty$, so, looking at ratios, that numerator might as well be $sqrt{n^2}$. Similarly, that denominator might as well be $sqrt[3]{n^3}$. Factoring out those big terms is a good way to deal with this (with algebraic functions.
Oh, and "much bigger" has a clear technical meaning here; the ratio goes to $infty$.
answered Jan 10 at 11:55
jmerryjmerry
17k11633
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
When using $ntoinfty$, it is generally assumed that we're dealing with a sequence; it doesn't really matter here, because the functions $xmapstosqrt{x^2+4x}$ and $xmapstosqrt[3]{x^3-3x^2}$ are monotonic in a neighborhood of $infty$.
L'Hôpital's theorem which you seem to refer to is one of the tools, but certainly not the tool of choice in every case. Indeed, if you try it, you're left with something like
$$
frac{x+2}{sqrt{x^2+4x}}frac{sqrt[3]{(x^3-3x^2)^2}}{x^2-2x}
$$
which is much worse than what you started with.
Another nice example is
$$
lim_{xtoinfty}frac{x}{sqrt{x^2+1}}overset{text{(H)}}{=}
lim_{xtoinfty}frac{1}{;dfrac{x}{sqrt{x^2+1}}}=
lim_{xtoinfty}frac{sqrt{x^2+1}}{x}
$$
and this loops.
In this case, collecting the largest power of $n$ is a good method; you can also do the substitution $n=1/t$ at the outset, getting
$$
lim_{tto0^+}frac{sqrt{dfrac{1}{t^2}+dfrac{4}{t}}}{sqrt[3]{dfrac{1}{t^3}-dfrac{3}{t^2}}}=
lim_{tto0^+}frac{dfrac{sqrt{1+4t}}{|t|}}{dfrac{sqrt[3]{1-3t}}{t}}=
lim_{tto0^+}frac{sqrt{1+4t}}{sqrt[3]{1-3t}}
$$
which is elementary.
answered Jan 10 at 12:12
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
When using $ntoinfty$, it is generally assumed that we're dealing with a sequence; it doesn't really matter here, because the functions $xmapstosqrt{x^2+4x}$ and $xmapstosqrt[3]{x^3-3x^2}$ are monotonic in a neighborhood of $infty$.
L'Hôpital's theorem which you seem to refer to is one of the tools, but certainly not the tool of choice in every case. Indeed, if you try it, you're left with something like
$$
frac{x+2}{sqrt{x^2+4x}}frac{sqrt[3]{(x^3-3x^2)^2}}{x^2-2x}
$$
which is much worse than what you started with.
Another nice example is
$$
lim_{xtoinfty}frac{x}{sqrt{x^2+1}}overset{text{(H)}}{=}
lim_{xtoinfty}frac{1}{;dfrac{x}{sqrt{x^2+1}}}=
lim_{xtoinfty}frac{sqrt{x^2+1}}{x}
$$
and this loops.
In this case, collecting the largest power of $n$ is a good method; you can also do the substitution $n=1/t$ at the outset, getting
$$
lim_{tto0^+}frac{sqrt{dfrac{1}{t^2}+dfrac{4}{t}}}{sqrt[3]{dfrac{1}{t^3}-dfrac{3}{t^2}}}=
lim_{tto0^+}frac{dfrac{sqrt{1+4t}}{|t|}}{dfrac{sqrt[3]{1-3t}}{t}}=
lim_{tto0^+}frac{sqrt{1+4t}}{sqrt[3]{1-3t}}
$$
which is elementary.
answered Jan 10 at 12:12
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
When using $ntoinfty$, it is generally assumed that we're dealing with a sequence; it doesn't really matter here, because the functions $xmapstosqrt{x^2+4x}$ and $xmapstosqrt[3]{x^3-3x^2}$ are monotonic in a neighborhood of $infty$.
L'Hôpital's theorem which you seem to refer to is one of the tools, but certainly not the tool of choice in every case. Indeed, if you try it, you're left with something like
$$
frac{x+2}{sqrt{x^2+4x}}frac{sqrt[3]{(x^3-3x^2)^2}}{x^2-2x}
$$
which is much worse than what you started with.
Another nice example is
$$
lim_{xtoinfty}frac{x}{sqrt{x^2+1}}overset{text{(H)}}{=}
lim_{xtoinfty}frac{1}{;dfrac{x}{sqrt{x^2+1}}}=
lim_{xtoinfty}frac{sqrt{x^2+1}}{x}
$$
and this loops.
In this case, collecting the largest power of $n$ is a good method; you can also do the substitution $n=1/t$ at the outset, getting
$$
lim_{tto0^+}frac{sqrt{dfrac{1}{t^2}+dfrac{4}{t}}}{sqrt[3]{dfrac{1}{t^3}-dfrac{3}{t^2}}}=
lim_{tto0^+}frac{dfrac{sqrt{1+4t}}{|t|}}{dfrac{sqrt[3]{1-3t}}{t}}=
lim_{tto0^+}frac{sqrt{1+4t}}{sqrt[3]{1-3t}}
$$
which is elementary.
answered Jan 10 at 12:12
egregegreg
185k1486208
$endgroup$
When using $ntoinfty$, it is generally assumed that we're dealing with a sequence; it doesn't really matter here, because the functions $xmapstosqrt{x^2+4x}$ and $xmapstosqrt[3]{x^3-3x^2}$ are monotonic in a neighborhood of $infty$.
L'Hôpital's theorem which you seem to refer to is one of the tools, but certainly not the tool of choice in every case. Indeed, if you try it, you're left with something like
$$
frac{x+2}{sqrt{x^2+4x}}frac{sqrt[3]{(x^3-3x^2)^2}}{x^2-2x}
$$
which is much worse than what you started with.
Another nice example is
$$
lim_{xtoinfty}frac{x}{sqrt{x^2+1}}overset{text{(H)}}{=}
lim_{xtoinfty}frac{1}{;dfrac{x}{sqrt{x^2+1}}}=
lim_{xtoinfty}frac{sqrt{x^2+1}}{x}
$$
and this loops.
In this case, collecting the largest power of $n$ is a good method; you can also do the substitution $n=1/t$ at the outset, getting
$$
lim_{tto0^+}frac{sqrt{dfrac{1}{t^2}+dfrac{4}{t}}}{sqrt[3]{dfrac{1}{t^3}-dfrac{3}{t^2}}}=
lim_{tto0^+}frac{dfrac{sqrt{1+4t}}{|t|}}{dfrac{sqrt[3]{1-3t}}{t}}=
lim_{tto0^+}frac{sqrt{1+4t}}{sqrt[3]{1-3t}}
$$
which is elementary.
answered Jan 10 at 12:12
egregegreg
185k1486208
answered Jan 10 at 12:12
egregegreg
185k1486208
answered Jan 10 at 12:12
egregegreg
185k1486208
answered Jan 10 at 12:12
egregegreg
185k1486208
185k1486208
add a comment |
add a comment |
$begingroup$
Well, I'd start by suggesting that you try to follow the solution printed there. We write $sqrt{n^2+text{stuff}}=nsqrt{1+text{small stuff}}$ and $sqrt[3]{n^3+text{stuff}}=nsqrt[3]{1+text{small stuff}}$. As $ntoinfty$, the parts I've labeled "small stuff" go to zero, and the limiting ratio is $frac{nsqrt{1}}{nsqrt[3]{1}}=frac nn=1$.
When dealing with limits, the first instinct should be to focus on what's big. $n^2$ is much bigger than $4n$ as $ntoinfty$, so, looking at ratios, that numerator might as well be $sqrt{n^2}$. Similarly, that denominator might as well be $sqrt[3]{n^3}$. Factoring out those big terms is a good way to deal with this (with algebraic functions.
Oh, and "much bigger" has a clear technical meaning here; the ratio goes to $infty$.
answered Jan 10 at 11:55
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
Well, I'd start by suggesting that you try to follow the solution printed there. We write $sqrt{n^2+text{stuff}}=nsqrt{1+text{small stuff}}$ and $sqrt[3]{n^3+text{stuff}}=nsqrt[3]{1+text{small stuff}}$. As $ntoinfty$, the parts I've labeled "small stuff" go to zero, and the limiting ratio is $frac{nsqrt{1}}{nsqrt[3]{1}}=frac nn=1$.
When dealing with limits, the first instinct should be to focus on what's big. $n^2$ is much bigger than $4n$ as $ntoinfty$, so, looking at ratios, that numerator might as well be $sqrt{n^2}$. Similarly, that denominator might as well be $sqrt[3]{n^3}$. Factoring out those big terms is a good way to deal with this (with algebraic functions.
Oh, and "much bigger" has a clear technical meaning here; the ratio goes to $infty$.
answered Jan 10 at 11:55
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
Well, I'd start by suggesting that you try to follow the solution printed there. We write $sqrt{n^2+text{stuff}}=nsqrt{1+text{small stuff}}$ and $sqrt[3]{n^3+text{stuff}}=nsqrt[3]{1+text{small stuff}}$. As $ntoinfty$, the parts I've labeled "small stuff" go to zero, and the limiting ratio is $frac{nsqrt{1}}{nsqrt[3]{1}}=frac nn=1$.
When dealing with limits, the first instinct should be to focus on what's big. $n^2$ is much bigger than $4n$ as $ntoinfty$, so, looking at ratios, that numerator might as well be $sqrt{n^2}$. Similarly, that denominator might as well be $sqrt[3]{n^3}$. Factoring out those big terms is a good way to deal with this (with algebraic functions.
Oh, and "much bigger" has a clear technical meaning here; the ratio goes to $infty$.
answered Jan 10 at 11:55
jmerryjmerry
17k11633
$endgroup$
Well, I'd start by suggesting that you try to follow the solution printed there. We write $sqrt{n^2+text{stuff}}=nsqrt{1+text{small stuff}}$ and $sqrt[3]{n^3+text{stuff}}=nsqrt[3]{1+text{small stuff}}$. As $ntoinfty$, the parts I've labeled "small stuff" go to zero, and the limiting ratio is $frac{nsqrt{1}}{nsqrt[3]{1}}=frac nn=1$.
When dealing with limits, the first instinct should be to focus on what's big. $n^2$ is much bigger than $4n$ as $ntoinfty$, so, looking at ratios, that numerator might as well be $sqrt{n^2}$. Similarly, that denominator might as well be $sqrt[3]{n^3}$. Factoring out those big terms is a good way to deal with this (with algebraic functions.
Oh, and "much bigger" has a clear technical meaning here; the ratio goes to $infty$.
answered Jan 10 at 11:55
jmerryjmerry
17k11633
answered Jan 10 at 11:55
jmerryjmerry
17k11633
answered Jan 10 at 11:55
jmerryjmerry
17k11633
answered Jan 10 at 11:55
jmerryjmerry
17k11633
17k11633
add a comment |
add a comment |
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2
$begingroup$
The photo is sidewise
$endgroup$
– William Elliot
Jan 10 at 11:32
$begingroup$
I know, what is the problem in it.
$endgroup$
– Akash
Jan 10 at 11:39
$begingroup$
I don't understand what is you question. You are given a solution below the exercise. What is going there? Well $n$ gets factored out and canceled, while then by continuity of root: $limsqrt{1+{a over n}} = sqrt{limleft({1 + {a over n}}right)} = sqrt{1 + lim{aover n}}$. What is the limit of ${aover n}$ as $n to infty$ given $a in Bbb R$ is some fixed number? ${aover n}$ gets infinitely small and hence the root is tending to $1$ as $n$ tends to infinity.
$endgroup$
– roman
Jan 10 at 11:54
$begingroup$
Also please use this reference on how to typeset math on this site. There are several examples of typesetting roots in there
$endgroup$
– roman
Jan 10 at 11:57