How to evaluate this limit: $lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac12$?












3












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How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.






calculus limits radicals







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  • 2




    $begingroup$
    Hint: Recall the definition of a derivative.
    $endgroup$
    – Andrew D
    Sep 10 '13 at 17:47












  • $begingroup$
    For some basic information about writing math at this site see e.g. here, here, here and here.
    $endgroup$
    – Martin Sleziak
    Sep 10 '13 at 17:50








  • 1




    $begingroup$
    The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
    $endgroup$
    – colormegone
    Sep 10 '13 at 17:52
















3












$begingroup$


How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.






calculus limits radicals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: Recall the definition of a derivative.
    $endgroup$
    – Andrew D
    Sep 10 '13 at 17:47












  • $begingroup$
    For some basic information about writing math at this site see e.g. here, here, here and here.
    $endgroup$
    – Martin Sleziak
    Sep 10 '13 at 17:50








  • 1




    $begingroup$
    The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
    $endgroup$
    – colormegone
    Sep 10 '13 at 17:52














3












3








3





$begingroup$


How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.






calculus limits radicals







share|cite|improve this question











$endgroup$




How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.






calculus limits radicals




calculus limits radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 5 '16 at 18:40









Martin Sleziak

44.7k9117272




44.7k9117272










asked Sep 10 '13 at 17:46









kb95825kb95825

47114




47114








  • 2




    $begingroup$
    Hint: Recall the definition of a derivative.
    $endgroup$
    – Andrew D
    Sep 10 '13 at 17:47












  • $begingroup$
    For some basic information about writing math at this site see e.g. here, here, here and here.
    $endgroup$
    – Martin Sleziak
    Sep 10 '13 at 17:50








  • 1




    $begingroup$
    The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
    $endgroup$
    – colormegone
    Sep 10 '13 at 17:52














  • 2




    $begingroup$
    Hint: Recall the definition of a derivative.
    $endgroup$
    – Andrew D
    Sep 10 '13 at 17:47












  • $begingroup$
    For some basic information about writing math at this site see e.g. here, here, here and here.
    $endgroup$
    – Martin Sleziak
    Sep 10 '13 at 17:50








  • 1




    $begingroup$
    The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
    $endgroup$
    – colormegone
    Sep 10 '13 at 17:52








2




2




$begingroup$
Hint: Recall the definition of a derivative.
$endgroup$
– Andrew D
Sep 10 '13 at 17:47






$begingroup$
Hint: Recall the definition of a derivative.
$endgroup$
– Andrew D
Sep 10 '13 at 17:47














$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Sep 10 '13 at 17:50






$begingroup$
For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Sep 10 '13 at 17:50






1




1




$begingroup$
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
$endgroup$
– colormegone
Sep 10 '13 at 17:52




$begingroup$
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
$endgroup$
– colormegone
Sep 10 '13 at 17:52










4 Answers
4






active

oldest

votes


















7












$begingroup$

$require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.



When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$



$$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$



Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$



I trust you can do that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
    $endgroup$
    – kb95825
    Sep 10 '13 at 17:50










  • $begingroup$
    Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
    $endgroup$
    – amWhy
    Sep 10 '13 at 17:51












  • $begingroup$
    @amWhy I suggest you fix the braces in the remainders ;-)
    $endgroup$
    – AlexR
    Sep 10 '13 at 17:57










  • $begingroup$
    @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
    $endgroup$
    – amWhy
    Sep 10 '13 at 17:57










  • $begingroup$
    yes thank you :)
    $endgroup$
    – kb95825
    Sep 10 '13 at 21:25



















2












$begingroup$

Method 1 (basic)

$$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)

$$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
where $f(y) = sqrt{y}$, thus the limit is
$$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
Method 3 (l'Hospital)

Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
$$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$






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  • $begingroup$
    Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
    $endgroup$
    – Martin Sleziak
    Sep 10 '13 at 17:54










  • $begingroup$
    @MartinSleziak Yep, that's true. I'll add it
    $endgroup$
    – AlexR
    Sep 10 '13 at 17:54



















0












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Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$






share|cite|improve this answer











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    0












    $begingroup$

    $x>-1$.



    $x= (x +1)- 1= sqrt{(x+1)}^2-1^2$



    $=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.



    $dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$



    Take the limit $x rightarrow 0$.






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      $require{cancel}$
      Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.



      When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$



      $$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$



      Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$



      I trust you can do that.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
        $endgroup$
        – kb95825
        Sep 10 '13 at 17:50










      • $begingroup$
        Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:51












      • $begingroup$
        @amWhy I suggest you fix the braces in the remainders ;-)
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:57










      • $begingroup$
        @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:57










      • $begingroup$
        yes thank you :)
        $endgroup$
        – kb95825
        Sep 10 '13 at 21:25
















      7












      $begingroup$

      $require{cancel}$
      Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.



      When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$



      $$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$



      Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$



      I trust you can do that.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
        $endgroup$
        – kb95825
        Sep 10 '13 at 17:50










      • $begingroup$
        Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:51












      • $begingroup$
        @amWhy I suggest you fix the braces in the remainders ;-)
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:57










      • $begingroup$
        @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:57










      • $begingroup$
        yes thank you :)
        $endgroup$
        – kb95825
        Sep 10 '13 at 21:25














      7












      7








      7





      $begingroup$

      $require{cancel}$
      Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.



      When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$



      $$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$



      Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$



      I trust you can do that.






      share|cite|improve this answer











      $endgroup$



      $require{cancel}$
      Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.



      When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$



      $$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$



      Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$



      I trust you can do that.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 9 '18 at 20:39

























      answered Sep 10 '13 at 17:47









      amWhyamWhy

      192k28225439




      192k28225439












      • $begingroup$
        what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
        $endgroup$
        – kb95825
        Sep 10 '13 at 17:50










      • $begingroup$
        Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:51












      • $begingroup$
        @amWhy I suggest you fix the braces in the remainders ;-)
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:57










      • $begingroup$
        @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:57










      • $begingroup$
        yes thank you :)
        $endgroup$
        – kb95825
        Sep 10 '13 at 21:25


















      • $begingroup$
        what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
        $endgroup$
        – kb95825
        Sep 10 '13 at 17:50










      • $begingroup$
        Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:51












      • $begingroup$
        @amWhy I suggest you fix the braces in the remainders ;-)
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:57










      • $begingroup$
        @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
        $endgroup$
        – amWhy
        Sep 10 '13 at 17:57










      • $begingroup$
        yes thank you :)
        $endgroup$
        – kb95825
        Sep 10 '13 at 21:25
















      $begingroup$
      what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
      $endgroup$
      – kb95825
      Sep 10 '13 at 17:50




      $begingroup$
      what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
      $endgroup$
      – kb95825
      Sep 10 '13 at 17:50












      $begingroup$
      Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
      $endgroup$
      – amWhy
      Sep 10 '13 at 17:51






      $begingroup$
      Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
      $endgroup$
      – amWhy
      Sep 10 '13 at 17:51














      $begingroup$
      @amWhy I suggest you fix the braces in the remainders ;-)
      $endgroup$
      – AlexR
      Sep 10 '13 at 17:57




      $begingroup$
      @amWhy I suggest you fix the braces in the remainders ;-)
      $endgroup$
      – AlexR
      Sep 10 '13 at 17:57












      $begingroup$
      @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
      $endgroup$
      – amWhy
      Sep 10 '13 at 17:57




      $begingroup$
      @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
      $endgroup$
      – amWhy
      Sep 10 '13 at 17:57












      $begingroup$
      yes thank you :)
      $endgroup$
      – kb95825
      Sep 10 '13 at 21:25




      $begingroup$
      yes thank you :)
      $endgroup$
      – kb95825
      Sep 10 '13 at 21:25











      2












      $begingroup$

      Method 1 (basic)

      $$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
      as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
      Method 2 (derivative)

      $$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
      where $f(y) = sqrt{y}$, thus the limit is
      $$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
      Method 3 (l'Hospital)

      Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
      $$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
        $endgroup$
        – Martin Sleziak
        Sep 10 '13 at 17:54










      • $begingroup$
        @MartinSleziak Yep, that's true. I'll add it
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:54
















      2












      $begingroup$

      Method 1 (basic)

      $$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
      as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
      Method 2 (derivative)

      $$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
      where $f(y) = sqrt{y}$, thus the limit is
      $$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
      Method 3 (l'Hospital)

      Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
      $$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
        $endgroup$
        – Martin Sleziak
        Sep 10 '13 at 17:54










      • $begingroup$
        @MartinSleziak Yep, that's true. I'll add it
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:54














      2












      2








      2





      $begingroup$

      Method 1 (basic)

      $$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
      as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
      Method 2 (derivative)

      $$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
      where $f(y) = sqrt{y}$, thus the limit is
      $$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
      Method 3 (l'Hospital)

      Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
      $$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$






      share|cite|improve this answer











      $endgroup$



      Method 1 (basic)

      $$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
      as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
      Method 2 (derivative)

      $$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
      where $f(y) = sqrt{y}$, thus the limit is
      $$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
      Method 3 (l'Hospital)

      Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
      $$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Sep 10 '13 at 17:58

























      answered Sep 10 '13 at 17:51









      AlexRAlexR

      22.7k12349




      22.7k12349












      • $begingroup$
        Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
        $endgroup$
        – Martin Sleziak
        Sep 10 '13 at 17:54










      • $begingroup$
        @MartinSleziak Yep, that's true. I'll add it
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:54


















      • $begingroup$
        Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
        $endgroup$
        – Martin Sleziak
        Sep 10 '13 at 17:54










      • $begingroup$
        @MartinSleziak Yep, that's true. I'll add it
        $endgroup$
        – AlexR
        Sep 10 '13 at 17:54
















      $begingroup$
      Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
      $endgroup$
      – Martin Sleziak
      Sep 10 '13 at 17:54




      $begingroup$
      Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
      $endgroup$
      – Martin Sleziak
      Sep 10 '13 at 17:54












      $begingroup$
      @MartinSleziak Yep, that's true. I'll add it
      $endgroup$
      – AlexR
      Sep 10 '13 at 17:54




      $begingroup$
      @MartinSleziak Yep, that's true. I'll add it
      $endgroup$
      – AlexR
      Sep 10 '13 at 17:54











      0












      $begingroup$

      Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$






          share|cite|improve this answer











          $endgroup$



          Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 21:36

























          answered Dec 9 '18 at 20:42









          greedoidgreedoid

          39.3k114797




          39.3k114797























              0












              $begingroup$

              $x>-1$.



              $x= (x +1)- 1= sqrt{(x+1)}^2-1^2$



              $=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.



              $dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$



              Take the limit $x rightarrow 0$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $x>-1$.



                $x= (x +1)- 1= sqrt{(x+1)}^2-1^2$



                $=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.



                $dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$



                Take the limit $x rightarrow 0$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $x>-1$.



                  $x= (x +1)- 1= sqrt{(x+1)}^2-1^2$



                  $=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.



                  $dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$



                  Take the limit $x rightarrow 0$.






                  share|cite|improve this answer









                  $endgroup$



                  $x>-1$.



                  $x= (x +1)- 1= sqrt{(x+1)}^2-1^2$



                  $=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.



                  $dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$



                  Take the limit $x rightarrow 0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 21:46









                  Peter SzilasPeter Szilas

                  11k2821




                  11k2821






























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