How to calculate $lim_{xto 0} frac{sin{(pi sqrt {x+1} )}}{x}$ without L'Hospital's rule? [closed]











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I got stuck trying to calculate the following limits:



$$
lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
$$



I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.










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closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

















    up vote
    -3
    down vote

    favorite
    1












    I got stuck trying to calculate the following limits:



    $$
    lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
    lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
    $$



    I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.










    share|cite|improve this question















    closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh

    If this question can be reworded to fit the rules in the help center, please edit the question.















      up vote
      -3
      down vote

      favorite
      1









      up vote
      -3
      down vote

      favorite
      1






      1





      I got stuck trying to calculate the following limits:



      $$
      lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
      lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
      $$



      I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.










      share|cite|improve this question















      I got stuck trying to calculate the following limits:



      $$
      lim_{xto 0-} frac{sin{(pi sqrt {x+1} )}}{x},quad
      lim_{xto 0+} frac{sin{(pi sqrt {x+1} )}}{x}
      $$



      I've tried to approximate the limit using the three functions method, but had no luck. It is also prohibited to use L'Hospital's rule.







      calculus limits limits-without-lhopital






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      share|cite|improve this question













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      edited Nov 15 at 19:42









      Bernard

      116k637108




      116k637108










      asked Nov 15 at 19:10









      absalon

      11




      11




      closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh Nov 26 at 0:32


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user302797, amWhy, José Carlos Santos, Shailesh

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






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          up vote
          3
          down vote



          accepted










          In both cases
          $$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$






          share|cite|improve this answer























          • Why does the second factor converge to pi*(-1/2)?
            – absalon
            Nov 15 at 19:26










          • @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
            – hamam_Abdallah
            Nov 15 at 19:30










          • $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
            – Nosrati
            Nov 15 at 19:31












          • When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
            – amWhy
            Nov 25 at 18:38


















          up vote
          0
          down vote













          As an alternative by $f(x)=sin (pisqrt{x+1})$ we have



          $$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$



          or by binomial first order expansion




          • $sqrt{x+1}=1+frac12x +o(x)$


          $$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$






          share|cite|improve this answer






























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            In both cases
            $$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$






            share|cite|improve this answer























            • Why does the second factor converge to pi*(-1/2)?
              – absalon
              Nov 15 at 19:26










            • @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
              – hamam_Abdallah
              Nov 15 at 19:30










            • $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
              – Nosrati
              Nov 15 at 19:31












            • When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
              – amWhy
              Nov 25 at 18:38















            up vote
            3
            down vote



            accepted










            In both cases
            $$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$






            share|cite|improve this answer























            • Why does the second factor converge to pi*(-1/2)?
              – absalon
              Nov 15 at 19:26










            • @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
              – hamam_Abdallah
              Nov 15 at 19:30










            • $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
              – Nosrati
              Nov 15 at 19:31












            • When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
              – amWhy
              Nov 25 at 18:38













            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            In both cases
            $$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$






            share|cite|improve this answer














            In both cases
            $$lim_{xto 0}frac{sinpisqrt{x+1}}{x}=lim_{xto 0}frac{sin(pi-pisqrt{x+1})}{pi-pisqrt{x+1}}frac{pi-pisqrt{x+1}}{x}=1cdotpicdot-dfrac12=color{blue}{-frac{pi}{2}}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 at 18:37









            amWhy

            191k27223439




            191k27223439










            answered Nov 15 at 19:15









            Nosrati

            26.2k62353




            26.2k62353












            • Why does the second factor converge to pi*(-1/2)?
              – absalon
              Nov 15 at 19:26










            • @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
              – hamam_Abdallah
              Nov 15 at 19:30










            • $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
              – Nosrati
              Nov 15 at 19:31












            • When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
              – amWhy
              Nov 25 at 18:38


















            • Why does the second factor converge to pi*(-1/2)?
              – absalon
              Nov 15 at 19:26










            • @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
              – hamam_Abdallah
              Nov 15 at 19:30










            • $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
              – Nosrati
              Nov 15 at 19:31












            • When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
              – amWhy
              Nov 25 at 18:38
















            Why does the second factor converge to pi*(-1/2)?
            – absalon
            Nov 15 at 19:26




            Why does the second factor converge to pi*(-1/2)?
            – absalon
            Nov 15 at 19:26












            @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
            – hamam_Abdallah
            Nov 15 at 19:30




            @absalon $1/2$ comes from the derivative of $sqrt{x+1}$.
            – hamam_Abdallah
            Nov 15 at 19:30












            $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
            – Nosrati
            Nov 15 at 19:31






            $$dfrac{1-sqrt{x+1}}{x}timesdfrac{1+sqrt{x+1}}{1+sqrt{x+1}}$$
            – Nosrati
            Nov 15 at 19:31














            When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
            – amWhy
            Nov 25 at 18:38




            When you are displaying a formula or equation, no need for "dfrac" (overkill), as in $$dfrac xy$$. The result can be had via $$frac xy$$
            – amWhy
            Nov 25 at 18:38










            up vote
            0
            down vote













            As an alternative by $f(x)=sin (pisqrt{x+1})$ we have



            $$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$



            or by binomial first order expansion




            • $sqrt{x+1}=1+frac12x +o(x)$


            $$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$






            share|cite|improve this answer



























              up vote
              0
              down vote













              As an alternative by $f(x)=sin (pisqrt{x+1})$ we have



              $$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$



              or by binomial first order expansion




              • $sqrt{x+1}=1+frac12x +o(x)$


              $$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                As an alternative by $f(x)=sin (pisqrt{x+1})$ we have



                $$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$



                or by binomial first order expansion




                • $sqrt{x+1}=1+frac12x +o(x)$


                $$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$






                share|cite|improve this answer














                As an alternative by $f(x)=sin (pisqrt{x+1})$ we have



                $$lim_{xto0}dfrac{sinpisqrt{x+1}}{x}=f’(0)=-frac{pi}2$$



                or by binomial first order expansion




                • $sqrt{x+1}=1+frac12x +o(x)$


                $$frac{sinpisqrt{x+1}}{x}=frac{sinleft(pi+frac{pi}2x +o(x)right)}{x}=-frac{sinleft(frac{pi}2x +o(x)right)}{frac{pi}2x +o(x)}frac{frac{pi}2x +o(x)}{x}to -frac{pi}2$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 15 at 20:06

























                answered Nov 15 at 19:59









                gimusi

                90.6k74495




                90.6k74495















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