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







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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















up vote
23
down vote

favorite
3












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







share|cite|improve this question




















  • 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













up vote
23
down vote

favorite
3









up vote
23
down vote

favorite
3






3





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







share|cite|improve this question















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






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edited Dec 12 at 4:44

























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














  • 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










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.






share|cite|improve this answer























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    active

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    active

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    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.






    share|cite|improve this answer



























      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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer














        Your mistake is in writing



        $$frac 1 x = sqrt{frac{1}{x^2}}.$$



        Since $x < 0$, the correct version includes a negative sign.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Dec 11 at 2:44


























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