Limit at negative infinity, get negative of correct 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 2 hours ago
Cdizzle
906
add a comment |
up vote
10
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 2 hours ago
Cdizzle
906
4
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
2 hours ago
add a comment |
up vote
10
down vote
favorite
up vote
10
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 2 hours ago
Cdizzle
906
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 2 hours ago
Cdizzle
906
asked 2 hours ago
Cdizzle
906
asked 2 hours ago
Cdizzle
906
asked 2 hours ago
Cdizzle
906
asked 2 hours ago
Cdizzle
906
906
4
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
2 hours ago
add a comment |
4
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
2 hours ago
4
4
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
2 hours ago
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
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
9
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 |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
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
9
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
9
down vote
accepted
up vote
9
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 2 hours ago
community wiki
T. Bongers
add a comment |
add a comment |
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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
2 hours ago