Use the definition of derivative and find the following limit:

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up vote
1
down vote

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I do not understand what this question is asking me to do.



What does it mean to get the limit at 0 and how does that relate to the derivative using this example?



Are not the limit and the derivative at 0 going to be different?



I am really confused as to how I need to approach this question, do I take the derivative of the limit at 0?



I am probably misinterpreting this question altogether, please help me clarify? Thank you.
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limits derivatives definition







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    up vote
    1
    down vote

    favorite
    2












    I do not understand what this question is asking me to do.



    What does it mean to get the limit at 0 and how does that relate to the derivative using this example?



    Are not the limit and the derivative at 0 going to be different?



    I am really confused as to how I need to approach this question, do I take the derivative of the limit at 0?



    I am probably misinterpreting this question altogether, please help me clarify? Thank you.
    enter image description here






    limits derivatives definition







    share|cite|improve this question























      up vote
      1
      down vote

      favorite
      2









      up vote
      1
      down vote

      favorite
      2






      2





      I do not understand what this question is asking me to do.



      What does it mean to get the limit at 0 and how does that relate to the derivative using this example?



      Are not the limit and the derivative at 0 going to be different?



      I am really confused as to how I need to approach this question, do I take the derivative of the limit at 0?



      I am probably misinterpreting this question altogether, please help me clarify? Thank you.
      enter image description here






      limits derivatives definition







      share|cite|improve this question













      I do not understand what this question is asking me to do.



      What does it mean to get the limit at 0 and how does that relate to the derivative using this example?



      Are not the limit and the derivative at 0 going to be different?



      I am really confused as to how I need to approach this question, do I take the derivative of the limit at 0?



      I am probably misinterpreting this question altogether, please help me clarify? Thank you.
      enter image description here






      limits derivatives definition




      limits derivatives definition






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked 1 hour ago









      Josh Teal

      538




      538




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote













          Hint:



          Use the following property, if $f$ is differentiable,



          $$lim_h to 0 fracf(y+mh) - f(y-nh)(m+n)h=f'(y)$$






          share|cite|improve this answer
















          • 1




            better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
            – Jimmy Sabater
            47 mins ago










          • nice approach. =)
            – Siong Thye Goh
            44 mins ago










          • Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
            – Jimmy Sabater
            42 mins ago







          • 1




            hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
            – Siong Thye Goh
            31 mins ago


















          up vote
          2
          down vote













          Notice that $ ln (2x + 1 ) - ln (1-3x) = ln left( frac2x+11-3x right )$. Let $f(x) = ln left( frac2x+11-3x right )$ and $f(0) = ln 1 = 0 $. Now, your limit reads as



          beginalign*
          lim_x to 0 dfracln (2x + 1 ) - ln (1-3x)x &= lim_x to 0 frac ln left( frac2x+11-3x right ) x \
          &=lim_x to 0 frac f(x) - f(0) x-0 \
          &= f'(0)
          endalign*



          Can you finish it??






          share|cite|improve this answer




















            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            Hint:



            Use the following property, if $f$ is differentiable,



            $$lim_h to 0 fracf(y+mh) - f(y-nh)(m+n)h=f'(y)$$






            share|cite|improve this answer
















            • 1




              better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
              – Jimmy Sabater
              47 mins ago










            • nice approach. =)
              – Siong Thye Goh
              44 mins ago










            • Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
              – Jimmy Sabater
              42 mins ago







            • 1




              hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
              – Siong Thye Goh
              31 mins ago















            up vote
            3
            down vote













            Hint:



            Use the following property, if $f$ is differentiable,



            $$lim_h to 0 fracf(y+mh) - f(y-nh)(m+n)h=f'(y)$$






            share|cite|improve this answer
















            • 1




              better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
              – Jimmy Sabater
              47 mins ago










            • nice approach. =)
              – Siong Thye Goh
              44 mins ago










            • Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
              – Jimmy Sabater
              42 mins ago







            • 1




              hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
              – Siong Thye Goh
              31 mins ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            Hint:



            Use the following property, if $f$ is differentiable,



            $$lim_h to 0 fracf(y+mh) - f(y-nh)(m+n)h=f'(y)$$






            share|cite|improve this answer












            Hint:



            Use the following property, if $f$ is differentiable,



            $$lim_h to 0 fracf(y+mh) - f(y-nh)(m+n)h=f'(y)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Siong Thye Goh

            89.9k1460112




            89.9k1460112







            • 1




              better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
              – Jimmy Sabater
              47 mins ago










            • nice approach. =)
              – Siong Thye Goh
              44 mins ago










            • Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
              – Jimmy Sabater
              42 mins ago







            • 1




              hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
              – Siong Thye Goh
              31 mins ago













            • 1




              better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
              – Jimmy Sabater
              47 mins ago










            • nice approach. =)
              – Siong Thye Goh
              44 mins ago










            • Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
              – Jimmy Sabater
              42 mins ago







            • 1




              hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
              – Siong Thye Goh
              31 mins ago








            1




            1




            better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
            – Jimmy Sabater
            47 mins ago




            better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1)
            – Jimmy Sabater
            47 mins ago












            nice approach. =)
            – Siong Thye Goh
            44 mins ago




            nice approach. =)
            – Siong Thye Goh
            44 mins ago












            Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
            – Jimmy Sabater
            42 mins ago





            Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book.
            – Jimmy Sabater
            42 mins ago





            1




            1




            hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
            – Siong Thye Goh
            31 mins ago





            hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago.
            – Siong Thye Goh
            31 mins ago











            up vote
            2
            down vote













            Notice that $ ln (2x + 1 ) - ln (1-3x) = ln left( frac2x+11-3x right )$. Let $f(x) = ln left( frac2x+11-3x right )$ and $f(0) = ln 1 = 0 $. Now, your limit reads as



            beginalign*
            lim_x to 0 dfracln (2x + 1 ) - ln (1-3x)x &= lim_x to 0 frac ln left( frac2x+11-3x right ) x \
            &=lim_x to 0 frac f(x) - f(0) x-0 \
            &= f'(0)
            endalign*



            Can you finish it??






            share|cite|improve this answer
























              up vote
              2
              down vote













              Notice that $ ln (2x + 1 ) - ln (1-3x) = ln left( frac2x+11-3x right )$. Let $f(x) = ln left( frac2x+11-3x right )$ and $f(0) = ln 1 = 0 $. Now, your limit reads as



              beginalign*
              lim_x to 0 dfracln (2x + 1 ) - ln (1-3x)x &= lim_x to 0 frac ln left( frac2x+11-3x right ) x \
              &=lim_x to 0 frac f(x) - f(0) x-0 \
              &= f'(0)
              endalign*



              Can you finish it??






              share|cite|improve this answer






















                up vote
                2
                down vote










                up vote
                2
                down vote









                Notice that $ ln (2x + 1 ) - ln (1-3x) = ln left( frac2x+11-3x right )$. Let $f(x) = ln left( frac2x+11-3x right )$ and $f(0) = ln 1 = 0 $. Now, your limit reads as



                beginalign*
                lim_x to 0 dfracln (2x + 1 ) - ln (1-3x)x &= lim_x to 0 frac ln left( frac2x+11-3x right ) x \
                &=lim_x to 0 frac f(x) - f(0) x-0 \
                &= f'(0)
                endalign*



                Can you finish it??






                share|cite|improve this answer












                Notice that $ ln (2x + 1 ) - ln (1-3x) = ln left( frac2x+11-3x right )$. Let $f(x) = ln left( frac2x+11-3x right )$ and $f(0) = ln 1 = 0 $. Now, your limit reads as



                beginalign*
                lim_x to 0 dfracln (2x + 1 ) - ln (1-3x)x &= lim_x to 0 frac ln left( frac2x+11-3x right ) x \
                &=lim_x to 0 frac f(x) - f(0) x-0 \
                &= f'(0)
                endalign*



                Can you finish it??







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 51 mins ago









                Jimmy Sabater

                1,827216




                1,827216



























                     

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