Evaluating $lim_{hto 0}frac{1}{h}(,(x+h)sec(x+h) - x sec(x),)$ without L'Hôpital's rule [closed]












0












$begingroup$


I got this problem as a high school problem. I want to find




$$lim_{hto 0}frac{(x+h)sec(x+h) - xsec(x)}{h}$$




I know that $$lim_{xto 0}frac{sin x}{x}=1qquadtext{and}qquadlim_{xto 0}frac{1 - cos x}{x}=0qquadtext{and}qquadlim_{xto 0}frac{tan x}{x}=1$$
I need to solve it using mainly these three standard forms.






limits trigonometry limits-without-lhopital







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closed as off-topic by Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber♦ Jan 8 at 23:13


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber

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
















  • $begingroup$
    Note that this is, by definition, $left(x sec xright)'$.
    $endgroup$
    – StackTD
    Jan 8 at 16:29










  • $begingroup$
    I want to find what work you did. Do you know the derivatives of sine and cosine? The $(sin h)/h$ limit as $h$ goes to zero?
    $endgroup$
    – Oscar Lanzi
    Jan 8 at 16:30












  • $begingroup$
    @OscarLanzi Yes I know
    $endgroup$
    – user630305
    Jan 8 at 16:43
















0












$begingroup$


I got this problem as a high school problem. I want to find




$$lim_{hto 0}frac{(x+h)sec(x+h) - xsec(x)}{h}$$




I know that $$lim_{xto 0}frac{sin x}{x}=1qquadtext{and}qquadlim_{xto 0}frac{1 - cos x}{x}=0qquadtext{and}qquadlim_{xto 0}frac{tan x}{x}=1$$
I need to solve it using mainly these three standard forms.






limits trigonometry limits-without-lhopital







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber♦ Jan 8 at 23:13


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber

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
















  • $begingroup$
    Note that this is, by definition, $left(x sec xright)'$.
    $endgroup$
    – StackTD
    Jan 8 at 16:29










  • $begingroup$
    I want to find what work you did. Do you know the derivatives of sine and cosine? The $(sin h)/h$ limit as $h$ goes to zero?
    $endgroup$
    – Oscar Lanzi
    Jan 8 at 16:30












  • $begingroup$
    @OscarLanzi Yes I know
    $endgroup$
    – user630305
    Jan 8 at 16:43














0












0








0





$begingroup$


I got this problem as a high school problem. I want to find




$$lim_{hto 0}frac{(x+h)sec(x+h) - xsec(x)}{h}$$




I know that $$lim_{xto 0}frac{sin x}{x}=1qquadtext{and}qquadlim_{xto 0}frac{1 - cos x}{x}=0qquadtext{and}qquadlim_{xto 0}frac{tan x}{x}=1$$
I need to solve it using mainly these three standard forms.






limits trigonometry limits-without-lhopital







share|cite|improve this question











$endgroup$




I got this problem as a high school problem. I want to find




$$lim_{hto 0}frac{(x+h)sec(x+h) - xsec(x)}{h}$$




I know that $$lim_{xto 0}frac{sin x}{x}=1qquadtext{and}qquadlim_{xto 0}frac{1 - cos x}{x}=0qquadtext{and}qquadlim_{xto 0}frac{tan x}{x}=1$$
I need to solve it using mainly these three standard forms.






limits trigonometry limits-without-lhopital




limits trigonometry limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 19:41









Blue

48k870153




48k870153










asked Jan 8 at 16:15









user630305user630305

63




63




closed as off-topic by Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber♦ Jan 8 at 23:13


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber

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







closed as off-topic by Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber♦ Jan 8 at 23:13


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, MisterRiemann, TravisJ, Siong Thye Goh, Alexander Gruber

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












  • $begingroup$
    Note that this is, by definition, $left(x sec xright)'$.
    $endgroup$
    – StackTD
    Jan 8 at 16:29










  • $begingroup$
    I want to find what work you did. Do you know the derivatives of sine and cosine? The $(sin h)/h$ limit as $h$ goes to zero?
    $endgroup$
    – Oscar Lanzi
    Jan 8 at 16:30












  • $begingroup$
    @OscarLanzi Yes I know
    $endgroup$
    – user630305
    Jan 8 at 16:43


















  • $begingroup$
    Note that this is, by definition, $left(x sec xright)'$.
    $endgroup$
    – StackTD
    Jan 8 at 16:29










  • $begingroup$
    I want to find what work you did. Do you know the derivatives of sine and cosine? The $(sin h)/h$ limit as $h$ goes to zero?
    $endgroup$
    – Oscar Lanzi
    Jan 8 at 16:30












  • $begingroup$
    @OscarLanzi Yes I know
    $endgroup$
    – user630305
    Jan 8 at 16:43
















$begingroup$
Note that this is, by definition, $left(x sec xright)'$.
$endgroup$
– StackTD
Jan 8 at 16:29




$begingroup$
Note that this is, by definition, $left(x sec xright)'$.
$endgroup$
– StackTD
Jan 8 at 16:29












$begingroup$
I want to find what work you did. Do you know the derivatives of sine and cosine? The $(sin h)/h$ limit as $h$ goes to zero?
$endgroup$
– Oscar Lanzi
Jan 8 at 16:30






$begingroup$
I want to find what work you did. Do you know the derivatives of sine and cosine? The $(sin h)/h$ limit as $h$ goes to zero?
$endgroup$
– Oscar Lanzi
Jan 8 at 16:30














$begingroup$
@OscarLanzi Yes I know
$endgroup$
– user630305
Jan 8 at 16:43




$begingroup$
@OscarLanzi Yes I know
$endgroup$
– user630305
Jan 8 at 16:43










2 Answers
2






active

oldest

votes


















1












$begingroup$

$$dfrac{(x+h)cos x-xcos(x+h)}{hcos xcos(x+h)}$$



$$=dfrac1{cos xcos(x+h)}left(xcdotdfrac{cos x-cos(x+h)}{h}+cos xright)$$



Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Maybe try something like this. :-)



    $$lim_{hto0}frac{(x+h)sec(x+h)-xsec x}{h}$$
    $$= lim_{hto0}frac{(x+h)sec(x+h)-xsec(x+h) +xsec(x+h)-xsec x}{h}$$






    share|cite|improve this answer











    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      $$dfrac{(x+h)cos x-xcos(x+h)}{hcos xcos(x+h)}$$



      $$=dfrac1{cos xcos(x+h)}left(xcdotdfrac{cos x-cos(x+h)}{h}+cos xright)$$



      Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $$dfrac{(x+h)cos x-xcos(x+h)}{hcos xcos(x+h)}$$



        $$=dfrac1{cos xcos(x+h)}left(xcdotdfrac{cos x-cos(x+h)}{h}+cos xright)$$



        Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $$dfrac{(x+h)cos x-xcos(x+h)}{hcos xcos(x+h)}$$



          $$=dfrac1{cos xcos(x+h)}left(xcdotdfrac{cos x-cos(x+h)}{h}+cos xright)$$



          Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html






          share|cite|improve this answer









          $endgroup$



          $$dfrac{(x+h)cos x-xcos(x+h)}{hcos xcos(x+h)}$$



          $$=dfrac1{cos xcos(x+h)}left(xcdotdfrac{cos x-cos(x+h)}{h}+cos xright)$$



          Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 17:31









          lab bhattacharjeelab bhattacharjee

          225k15156274




          225k15156274























              1












              $begingroup$

              Maybe try something like this. :-)



              $$lim_{hto0}frac{(x+h)sec(x+h)-xsec x}{h}$$
              $$= lim_{hto0}frac{(x+h)sec(x+h)-xsec(x+h) +xsec(x+h)-xsec x}{h}$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Maybe try something like this. :-)



                $$lim_{hto0}frac{(x+h)sec(x+h)-xsec x}{h}$$
                $$= lim_{hto0}frac{(x+h)sec(x+h)-xsec(x+h) +xsec(x+h)-xsec x}{h}$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Maybe try something like this. :-)



                  $$lim_{hto0}frac{(x+h)sec(x+h)-xsec x}{h}$$
                  $$= lim_{hto0}frac{(x+h)sec(x+h)-xsec(x+h) +xsec(x+h)-xsec x}{h}$$






                  share|cite|improve this answer











                  $endgroup$



                  Maybe try something like this. :-)



                  $$lim_{hto0}frac{(x+h)sec(x+h)-xsec x}{h}$$
                  $$= lim_{hto0}frac{(x+h)sec(x+h)-xsec(x+h) +xsec(x+h)-xsec x}{h}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 8 at 18:26

























                  answered Jan 8 at 16:30









                  John JoyJohn Joy

                  6,20611526




                  6,20611526















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