Why does trying to compute $lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}}$ result in the negative of the answer...
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My textbook asks me to evaluate the limit $$lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}}$$ which evaluates to $-2oversqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} {2x-1over sqrt{x^2left(3+frac{1}{x}+frac{1}{x^2}right)}} \
& = lim_{xto-infty} {2x-1over -xsqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = lim_{xto-infty} {-2+frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {-2oversqrt{3}}
end{align}$$
the second step is justified because $xto-infty$ implies $xlt0$, so $sqrt{x^2}=-x$.
For my attempt I ended up with the negative of the correct answer:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} left({2x-1over sqrt{3x^2+x+1}}cdotfrac{frac{1}{x}}{frac{1}{x}}right) \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{frac{1}{x^2}left(3x^2+x+1right)}} \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {2oversqrt{3}}
end{align}$$
Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.
calculus algebra-precalculus limits
asked Dec 11 at 2:41
Cdizzle
1586
add a comment |
up vote
23
down vote
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My textbook asks me to evaluate the limit $$lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}}$$ which evaluates to $-2oversqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} {2x-1over sqrt{x^2left(3+frac{1}{x}+frac{1}{x^2}right)}} \
& = lim_{xto-infty} {2x-1over -xsqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = lim_{xto-infty} {-2+frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {-2oversqrt{3}}
end{align}$$
the second step is justified because $xto-infty$ implies $xlt0$, so $sqrt{x^2}=-x$.
For my attempt I ended up with the negative of the correct answer:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} left({2x-1over sqrt{3x^2+x+1}}cdotfrac{frac{1}{x}}{frac{1}{x}}right) \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{frac{1}{x^2}left(3x^2+x+1right)}} \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {2oversqrt{3}}
end{align}$$
Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.
calculus algebra-precalculus limits
asked Dec 11 at 2:41
Cdizzle
1586
17
Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question.
– T. Bongers
Dec 11 at 2:46
2
It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} ne (-1)^{6·1/2}$.
– user21820
Dec 11 at 9:08
In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis.
– Gautam Shenoy
Dec 11 at 14:55
1
Note that $displaystyle,sqrt{,{a^{2}},}, = leftvert,{a},rightvert$
– Felix Marin
Dec 12 at 5:33
add a comment |
up vote
23
down vote
favorite
up vote
23
down vote
favorite
My textbook asks me to evaluate the limit $$lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}}$$ which evaluates to $-2oversqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} {2x-1over sqrt{x^2left(3+frac{1}{x}+frac{1}{x^2}right)}} \
& = lim_{xto-infty} {2x-1over -xsqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = lim_{xto-infty} {-2+frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {-2oversqrt{3}}
end{align}$$
the second step is justified because $xto-infty$ implies $xlt0$, so $sqrt{x^2}=-x$.
For my attempt I ended up with the negative of the correct answer:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} left({2x-1over sqrt{3x^2+x+1}}cdotfrac{frac{1}{x}}{frac{1}{x}}right) \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{frac{1}{x^2}left(3x^2+x+1right)}} \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {2oversqrt{3}}
end{align}$$
Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.
calculus algebra-precalculus limits
asked Dec 11 at 2:41
Cdizzle
1586
My textbook asks me to evaluate the limit $$lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}}$$ which evaluates to $-2oversqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} {2x-1over sqrt{x^2left(3+frac{1}{x}+frac{1}{x^2}right)}} \
& = lim_{xto-infty} {2x-1over -xsqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = lim_{xto-infty} {-2+frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {-2oversqrt{3}}
end{align}$$
the second step is justified because $xto-infty$ implies $xlt0$, so $sqrt{x^2}=-x$.
For my attempt I ended up with the negative of the correct answer:
$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} left({2x-1over sqrt{3x^2+x+1}}cdotfrac{frac{1}{x}}{frac{1}{x}}right) \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{frac{1}{x^2}left(3x^2+x+1right)}} \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {2oversqrt{3}}
end{align}$$
Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.
calculus algebra-precalculus limits
calculus algebra-precalculus limits
asked Dec 11 at 2:41
Cdizzle
1586
asked Dec 11 at 2:41
Cdizzle
1586
edited Dec 12 at 4:44
asked Dec 11 at 2:41
Cdizzle
1586
asked Dec 11 at 2:41
Cdizzle
1586
asked Dec 11 at 2:41
Cdizzle
1586
1586
17
Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question.
– T. Bongers
Dec 11 at 2:46
2
It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} ne (-1)^{6·1/2}$.
– user21820
Dec 11 at 9:08
In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis.
– Gautam Shenoy
Dec 11 at 14:55
1
Note that $displaystyle,sqrt{,{a^{2}},}, = leftvert,{a},rightvert$
– Felix Marin
Dec 12 at 5:33
add a comment |
17
Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question.
– T. Bongers
Dec 11 at 2:46
2
It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} ne (-1)^{6·1/2}$.
– user21820
Dec 11 at 9:08
In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis.
– Gautam Shenoy
Dec 11 at 14:55
1
Note that $displaystyle,sqrt{,{a^{2}},}, = leftvert,{a},rightvert$
– Felix Marin
Dec 12 at 5:33
17
17
Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question.
– T. Bongers
Dec 11 at 2:46
Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question.
– T. Bongers
Dec 11 at 2:46
2
2
It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} ne (-1)^{6·1/2}$.
– user21820
Dec 11 at 9:08
It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} ne (-1)^{6·1/2}$.
– user21820
Dec 11 at 9:08
In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis.
– Gautam Shenoy
Dec 11 at 14:55
In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis.
– Gautam Shenoy
Dec 11 at 14:55
1
1
Note that $displaystyle,sqrt{,{a^{2}},}, = leftvert,{a},rightvert$
– Felix Marin
Dec 12 at 5:33
Note that $displaystyle,sqrt{,{a^{2}},}, = leftvert,{a},rightvert$
– Felix Marin
Dec 12 at 5:33
add a comment |
1 Answer
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Your mistake is in writing
$$frac 1 x = sqrt{frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
24
down vote
accepted
Your mistake is in writing
$$frac 1 x = sqrt{frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
add a comment |
up vote
24
down vote
accepted
Your mistake is in writing
$$frac 1 x = sqrt{frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
add a comment |
up vote
24
down vote
accepted
up vote
24
down vote
accepted
Your mistake is in writing
$$frac 1 x = sqrt{frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
Your mistake is in writing
$$frac 1 x = sqrt{frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
answered Dec 11 at 2:44
community wiki
T. Bongers
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17
Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question.
– T. Bongers
Dec 11 at 2:46
2
It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} ne (-1)^{6·1/2}$.
– user21820
Dec 11 at 9:08
In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis.
– Gautam Shenoy
Dec 11 at 14:55
1
Note that $displaystyle,sqrt{,{a^{2}},}, = leftvert,{a},rightvert$
– Felix Marin
Dec 12 at 5:33