How to calculate $lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$












0












$begingroup$


$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$



What I tried to do was to divide every term my $x^2$:



$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$



Then I calculated the limits of the numerator and denominator separately, which gave:



$$frac{0}{3}$$



For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?






limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:00










  • $begingroup$
    @T.Bongers not really, care to explain?
    $endgroup$
    – Trey
    Dec 3 '18 at 0:01










  • $begingroup$
    What is $sqrt{x^2}$?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:03










  • $begingroup$
    (1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
    $endgroup$
    – GEdgar
    Dec 3 '18 at 0:11
















0












$begingroup$


$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$



What I tried to do was to divide every term my $x^2$:



$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$



Then I calculated the limits of the numerator and denominator separately, which gave:



$$frac{0}{3}$$



For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?






limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:00










  • $begingroup$
    @T.Bongers not really, care to explain?
    $endgroup$
    – Trey
    Dec 3 '18 at 0:01










  • $begingroup$
    What is $sqrt{x^2}$?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:03










  • $begingroup$
    (1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
    $endgroup$
    – GEdgar
    Dec 3 '18 at 0:11














0












0








0





$begingroup$


$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$



What I tried to do was to divide every term my $x^2$:



$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$



Then I calculated the limits of the numerator and denominator separately, which gave:



$$frac{0}{3}$$



For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?






limits







share|cite|improve this question









$endgroup$




$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}$$



What I tried to do was to divide every term my $x^2$:



$$frac{frac{1}{x} + frac{2}{x^2}}{sqrt{9+frac{1}{x^2}}}$$



Then I calculated the limits of the numerator and denominator separately, which gave:



$$frac{0}{3}$$



For some reason though, it appears that the right answer is $frac{1}{3}$? Can someone explain me what's wrong with my solution?






limits




limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 23:59









TreyTrey

309113




309113












  • $begingroup$
    You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:00










  • $begingroup$
    @T.Bongers not really, care to explain?
    $endgroup$
    – Trey
    Dec 3 '18 at 0:01










  • $begingroup$
    What is $sqrt{x^2}$?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:03










  • $begingroup$
    (1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
    $endgroup$
    – GEdgar
    Dec 3 '18 at 0:11


















  • $begingroup$
    You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:00










  • $begingroup$
    @T.Bongers not really, care to explain?
    $endgroup$
    – Trey
    Dec 3 '18 at 0:01










  • $begingroup$
    What is $sqrt{x^2}$?
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:03










  • $begingroup$
    (1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
    $endgroup$
    – GEdgar
    Dec 3 '18 at 0:11
















$begingroup$
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
$endgroup$
– T. Bongers
Dec 3 '18 at 0:00




$begingroup$
You divided the numerator by $x^2$, but you really divided the denominator by $x^4$. Do you see why?
$endgroup$
– T. Bongers
Dec 3 '18 at 0:00












$begingroup$
@T.Bongers not really, care to explain?
$endgroup$
– Trey
Dec 3 '18 at 0:01




$begingroup$
@T.Bongers not really, care to explain?
$endgroup$
– Trey
Dec 3 '18 at 0:01












$begingroup$
What is $sqrt{x^2}$?
$endgroup$
– T. Bongers
Dec 3 '18 at 0:03




$begingroup$
What is $sqrt{x^2}$?
$endgroup$
– T. Bongers
Dec 3 '18 at 0:03












$begingroup$
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
$endgroup$
– GEdgar
Dec 3 '18 at 0:11




$begingroup$
(1) It is positive for lage $x$. (2) The limit of the square is easy to compute.
$endgroup$
– GEdgar
Dec 3 '18 at 0:11










2 Answers
2






active

oldest

votes


















3












$begingroup$

Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.



However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:



$$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:08










  • $begingroup$
    That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
    $endgroup$
    – gimusi
    Dec 3 '18 at 0:09












  • $begingroup$
    That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
    $endgroup$
    – Ben W
    Dec 3 '18 at 0:11












  • $begingroup$
    I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
    $endgroup$
    – gimusi
    Dec 3 '18 at 0:27










  • $begingroup$
    What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
    $endgroup$
    – Ben W
    Dec 3 '18 at 0:34





















1












$begingroup$

HINT



We have



$$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.



    However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:



    $$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
      $endgroup$
      – T. Bongers
      Dec 3 '18 at 0:08










    • $begingroup$
      That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:09












    • $begingroup$
      That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:11












    • $begingroup$
      I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:27










    • $begingroup$
      What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:34


















    3












    $begingroup$

    Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.



    However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:



    $$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
      $endgroup$
      – T. Bongers
      Dec 3 '18 at 0:08










    • $begingroup$
      That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:09












    • $begingroup$
      That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:11












    • $begingroup$
      I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:27










    • $begingroup$
      What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:34
















    3












    3








    3





    $begingroup$

    Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.



    However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:



    $$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$






    share|cite|improve this answer











    $endgroup$



    Your mistake is that when $1/x^2$ passes through a square root it becomes $sqrt{1/x^4}$. You should instead have hit numerator and denominator with $1/x$, not $1/x^2$.



    However, only leading terms matter. As a shortcut, it is perfectly fine (and much quicker) to reason as follows:



    $$lim_{xtoinfty}frac{x+2}{sqrt{9x^2+1}}=lim_{xtoinfty}frac{x}{sqrt{9x^2}}=frac{1}{3}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 0:14

























    answered Dec 3 '18 at 0:07









    Ben WBen W

    2,189615




    2,189615












    • $begingroup$
      Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
      $endgroup$
      – T. Bongers
      Dec 3 '18 at 0:08










    • $begingroup$
      That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:09












    • $begingroup$
      That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:11












    • $begingroup$
      I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:27










    • $begingroup$
      What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:34




















    • $begingroup$
      Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
      $endgroup$
      – T. Bongers
      Dec 3 '18 at 0:08










    • $begingroup$
      That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:09












    • $begingroup$
      That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:11












    • $begingroup$
      I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
      $endgroup$
      – gimusi
      Dec 3 '18 at 0:27










    • $begingroup$
      What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
      $endgroup$
      – Ben W
      Dec 3 '18 at 0:34


















    $begingroup$
    Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:08




    $begingroup$
    Although you're right in principle, I think it's worth noting that many of the people who are grading student work at this level would disagree that this is "perfectly fine" without more justification.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 0:08












    $begingroup$
    That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
    $endgroup$
    – gimusi
    Dec 3 '18 at 0:09






    $begingroup$
    That's a nice suggestion even if not completely rigorous, in that case I would prefer use the following notation $$frac{x+2}{sqrt{9x^2+1}} sim frac{x}{sqrt{9x^2}}to frac{1}{3}$$
    $endgroup$
    – gimusi
    Dec 3 '18 at 0:09














    $begingroup$
    That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
    $endgroup$
    – Ben W
    Dec 3 '18 at 0:11






    $begingroup$
    That notation is fine and I use it myself sometimes. However, I do not believe my suggestion is any less rigorous than any other. It simply presses into service different theorems than that $lim1/x^n=0$.
    $endgroup$
    – Ben W
    Dec 3 '18 at 0:11














    $begingroup$
    I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
    $endgroup$
    – gimusi
    Dec 3 '18 at 0:27




    $begingroup$
    I suppose you are appling that as equivalents but sometimes equivalents can induce to some mistakes. We need to be careful with that. Of course one expert user csn handle that without problem but a newbie must pat great attention with that.
    $endgroup$
    – gimusi
    Dec 3 '18 at 0:27












    $begingroup$
    What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
    $endgroup$
    – Ben W
    Dec 3 '18 at 0:34






    $begingroup$
    What I was applying was this fact: Let's say that a function $f(x)$ is polynomial-like if it has the form $f(x)=sum_{i=0}^na_i^{(f)}x^{b_i^{(f)}}$ for real coefficients $a_i^{(f)}$ and an increasing sequence $(b_i^{(f)})_{i=0}^n$ of nonnegative real numbers. Let $r$ and $s$ be positive real numbers, and let $g$ be another polynomial-like function. Then $lim_{xtoinfty}frac{[f(x)]^r}{[g(x)]^s}=lim_{xtoinfty}frac{[a_n^{(f)}x^{b_n^{(f)}]^r}{[a_n^{(g)}x^{b_n^{(g)}]^s}$.
    $endgroup$
    – Ben W
    Dec 3 '18 at 0:34













    1












    $begingroup$

    HINT



    We have



    $$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      HINT



      We have



      $$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        HINT



        We have



        $$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$






        share|cite|improve this answer









        $endgroup$



        HINT



        We have



        $$lim_{xto infty} frac{x+2}{sqrt{9x^2+1}}=lim_{xto infty} frac x{sqrt{x^2}}frac{1+2/x}{sqrt{9+1/x^2}}$$







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        answered Dec 3 '18 at 0:06









        gimusigimusi

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