Prove by definition $lim_ntoinfty sqrtfracnn+1=1$

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Prove by definition $lim_ntoinfty sqrtfracnn+1=1$.




I got to $left|sqrtfracnn+1-1right|< ε$, but I don't know how to proceed.




calculus limits




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    Prove by definition $lim_ntoinfty sqrtfracnn+1=1$.




    I got to $left|sqrtfracnn+1-1right|< ε$, but I don't know how to proceed.




    calculus limits




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Prove by definition $lim_ntoinfty sqrtfracnn+1=1$.




      I got to $left|sqrtfracnn+1-1right|< ε$, but I don't know how to proceed.




      calculus limits




      share|cite|improve this question














      Prove by definition $lim_ntoinfty sqrtfracnn+1=1$.




      I got to $left|sqrtfracnn+1-1right|< ε$, but I don't know how to proceed.





      calculus limits





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 31 at 19:47









      Math Lover

      12.2k21132




      12.2k21132









      asked Jul 31 at 19:43









      Matías López

      162




      162




















          2 Answers
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          $$beginarrayrcl
          left| sqrtdfrac n n+1 - 1 right|
          &=& left| dfrac sqrt n - sqrtn+1 sqrtn+1 right| \
          &=& left| dfrac 1 sqrtn+1 (sqrt n + sqrt n+1) right| \
          &le& left| dfrac 1 sqrtn (sqrt n + sqrt n) right| \
          &=& left| dfrac 1 2n right| \
          endarray$$






          share|cite|improve this answer





















          • Could you explain please?
            – Matías López
            Jul 31 at 20:03


















          up vote
          1
          down vote













          Let me explain Kenny Lau's answer a bit. The first step, he does



          $$left|sqrtfracnn+1-1right|=left|fracsqrtnsqrtn+1-1right|=left|fracsqrtnsqrtn+1-fracsqrtn+1sqrtn+1right|=left|fracsqrtn-sqrtn+1sqrtn+1right|$$



          Now he proceeds to multiply both sides of the fraction with $sqrtn+sqrtn+1$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes



          $$(sqrtn-sqrtn+1)(sqrtn+sqrtn+1)=sqrtn^2-sqrtn+1^2=n-(n+1)=-1$$



          so that we get



          $$left|fracsqrtn-sqrtn+1sqrtn+1right|=left|frac-1sqrtn+1(sqrtn+sqrtn+1)right|$$



          He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $sqrtn<sqrtn+1$ so that



          $$sqrtn+1(sqrtn+sqrtn+1)>sqrtn(sqrtn+sqrtn)=2n$$



          hence,



          $$frac1sqrtn+1(sqrtn+sqrtn+1)<frac1sqrtn(sqrtn+sqrtn)=frac12n$$



          so in the end



          $$left|sqrtfracnn+1-1right|<left|frac12nright|$$



          the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $epsilon$, because you can just make $frac12n$ smaller and then



          $$left|sqrtfracnn+1-1right|$$



          must be smaller too.






          share|cite|improve this answer





















          • Thanks! got it.
            – Matías López
            Aug 1 at 0:32










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          2 Answers
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          up vote
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          $$beginarrayrcl
          left| sqrtdfrac n n+1 - 1 right|
          &=& left| dfrac sqrt n - sqrtn+1 sqrtn+1 right| \
          &=& left| dfrac 1 sqrtn+1 (sqrt n + sqrt n+1) right| \
          &le& left| dfrac 1 sqrtn (sqrt n + sqrt n) right| \
          &=& left| dfrac 1 2n right| \
          endarray$$






          share|cite|improve this answer





















          • Could you explain please?
            – Matías López
            Jul 31 at 20:03















          up vote
          6
          down vote













          $$beginarrayrcl
          left| sqrtdfrac n n+1 - 1 right|
          &=& left| dfrac sqrt n - sqrtn+1 sqrtn+1 right| \
          &=& left| dfrac 1 sqrtn+1 (sqrt n + sqrt n+1) right| \
          &le& left| dfrac 1 sqrtn (sqrt n + sqrt n) right| \
          &=& left| dfrac 1 2n right| \
          endarray$$






          share|cite|improve this answer





















          • Could you explain please?
            – Matías López
            Jul 31 at 20:03













          up vote
          6
          down vote










          up vote
          6
          down vote









          $$beginarrayrcl
          left| sqrtdfrac n n+1 - 1 right|
          &=& left| dfrac sqrt n - sqrtn+1 sqrtn+1 right| \
          &=& left| dfrac 1 sqrtn+1 (sqrt n + sqrt n+1) right| \
          &le& left| dfrac 1 sqrtn (sqrt n + sqrt n) right| \
          &=& left| dfrac 1 2n right| \
          endarray$$






          share|cite|improve this answer













          $$beginarrayrcl
          left| sqrtdfrac n n+1 - 1 right|
          &=& left| dfrac sqrt n - sqrtn+1 sqrtn+1 right| \
          &=& left| dfrac 1 sqrtn+1 (sqrt n + sqrt n+1) right| \
          &le& left| dfrac 1 sqrtn (sqrt n + sqrt n) right| \
          &=& left| dfrac 1 2n right| \
          endarray$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 31 at 19:46









          Kenny Lau

          17.7k2156




          17.7k2156











          • Could you explain please?
            – Matías López
            Jul 31 at 20:03

















          • Could you explain please?
            – Matías López
            Jul 31 at 20:03
















          Could you explain please?
          – Matías López
          Jul 31 at 20:03





          Could you explain please?
          – Matías López
          Jul 31 at 20:03











          up vote
          1
          down vote













          Let me explain Kenny Lau's answer a bit. The first step, he does



          $$left|sqrtfracnn+1-1right|=left|fracsqrtnsqrtn+1-1right|=left|fracsqrtnsqrtn+1-fracsqrtn+1sqrtn+1right|=left|fracsqrtn-sqrtn+1sqrtn+1right|$$



          Now he proceeds to multiply both sides of the fraction with $sqrtn+sqrtn+1$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes



          $$(sqrtn-sqrtn+1)(sqrtn+sqrtn+1)=sqrtn^2-sqrtn+1^2=n-(n+1)=-1$$



          so that we get



          $$left|fracsqrtn-sqrtn+1sqrtn+1right|=left|frac-1sqrtn+1(sqrtn+sqrtn+1)right|$$



          He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $sqrtn<sqrtn+1$ so that



          $$sqrtn+1(sqrtn+sqrtn+1)>sqrtn(sqrtn+sqrtn)=2n$$



          hence,



          $$frac1sqrtn+1(sqrtn+sqrtn+1)<frac1sqrtn(sqrtn+sqrtn)=frac12n$$



          so in the end



          $$left|sqrtfracnn+1-1right|<left|frac12nright|$$



          the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $epsilon$, because you can just make $frac12n$ smaller and then



          $$left|sqrtfracnn+1-1right|$$



          must be smaller too.






          share|cite|improve this answer





















          • Thanks! got it.
            – Matías López
            Aug 1 at 0:32














          up vote
          1
          down vote













          Let me explain Kenny Lau's answer a bit. The first step, he does



          $$left|sqrtfracnn+1-1right|=left|fracsqrtnsqrtn+1-1right|=left|fracsqrtnsqrtn+1-fracsqrtn+1sqrtn+1right|=left|fracsqrtn-sqrtn+1sqrtn+1right|$$



          Now he proceeds to multiply both sides of the fraction with $sqrtn+sqrtn+1$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes



          $$(sqrtn-sqrtn+1)(sqrtn+sqrtn+1)=sqrtn^2-sqrtn+1^2=n-(n+1)=-1$$



          so that we get



          $$left|fracsqrtn-sqrtn+1sqrtn+1right|=left|frac-1sqrtn+1(sqrtn+sqrtn+1)right|$$



          He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $sqrtn<sqrtn+1$ so that



          $$sqrtn+1(sqrtn+sqrtn+1)>sqrtn(sqrtn+sqrtn)=2n$$



          hence,



          $$frac1sqrtn+1(sqrtn+sqrtn+1)<frac1sqrtn(sqrtn+sqrtn)=frac12n$$



          so in the end



          $$left|sqrtfracnn+1-1right|<left|frac12nright|$$



          the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $epsilon$, because you can just make $frac12n$ smaller and then



          $$left|sqrtfracnn+1-1right|$$



          must be smaller too.






          share|cite|improve this answer





















          • Thanks! got it.
            – Matías López
            Aug 1 at 0:32












          up vote
          1
          down vote










          up vote
          1
          down vote









          Let me explain Kenny Lau's answer a bit. The first step, he does



          $$left|sqrtfracnn+1-1right|=left|fracsqrtnsqrtn+1-1right|=left|fracsqrtnsqrtn+1-fracsqrtn+1sqrtn+1right|=left|fracsqrtn-sqrtn+1sqrtn+1right|$$



          Now he proceeds to multiply both sides of the fraction with $sqrtn+sqrtn+1$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes



          $$(sqrtn-sqrtn+1)(sqrtn+sqrtn+1)=sqrtn^2-sqrtn+1^2=n-(n+1)=-1$$



          so that we get



          $$left|fracsqrtn-sqrtn+1sqrtn+1right|=left|frac-1sqrtn+1(sqrtn+sqrtn+1)right|$$



          He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $sqrtn<sqrtn+1$ so that



          $$sqrtn+1(sqrtn+sqrtn+1)>sqrtn(sqrtn+sqrtn)=2n$$



          hence,



          $$frac1sqrtn+1(sqrtn+sqrtn+1)<frac1sqrtn(sqrtn+sqrtn)=frac12n$$



          so in the end



          $$left|sqrtfracnn+1-1right|<left|frac12nright|$$



          the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $epsilon$, because you can just make $frac12n$ smaller and then



          $$left|sqrtfracnn+1-1right|$$



          must be smaller too.






          share|cite|improve this answer













          Let me explain Kenny Lau's answer a bit. The first step, he does



          $$left|sqrtfracnn+1-1right|=left|fracsqrtnsqrtn+1-1right|=left|fracsqrtnsqrtn+1-fracsqrtn+1sqrtn+1right|=left|fracsqrtn-sqrtn+1sqrtn+1right|$$



          Now he proceeds to multiply both sides of the fraction with $sqrtn+sqrtn+1$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes



          $$(sqrtn-sqrtn+1)(sqrtn+sqrtn+1)=sqrtn^2-sqrtn+1^2=n-(n+1)=-1$$



          so that we get



          $$left|fracsqrtn-sqrtn+1sqrtn+1right|=left|frac-1sqrtn+1(sqrtn+sqrtn+1)right|$$



          He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $sqrtn<sqrtn+1$ so that



          $$sqrtn+1(sqrtn+sqrtn+1)>sqrtn(sqrtn+sqrtn)=2n$$



          hence,



          $$frac1sqrtn+1(sqrtn+sqrtn+1)<frac1sqrtn(sqrtn+sqrtn)=frac12n$$



          so in the end



          $$left|sqrtfracnn+1-1right|<left|frac12nright|$$



          the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $epsilon$, because you can just make $frac12n$ smaller and then



          $$left|sqrtfracnn+1-1right|$$



          must be smaller too.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 31 at 20:56









          vrugtehagel

          10.3k1548




          10.3k1548











          • Thanks! got it.
            – Matías López
            Aug 1 at 0:32
















          • Thanks! got it.
            – Matías López
            Aug 1 at 0:32















          Thanks! got it.
          – Matías López
          Aug 1 at 0:32




          Thanks! got it.
          – Matías López
          Aug 1 at 0:32












           

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