Limit addition and multiplication question












1















As x approaches one, the limit of f of x is 6, the limit of g of x is
8, and the limit of h of x is 10. What is the limit as x approaches
one of the function f + g times h?




So WLOG I just set $f(x) = 6/x$, $g(x) = 8/x$ and $h(x) = 10/x$.



Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$



So $lim_{xto 1} 140/x^{2} = 140$. So my answer is $140$



But the correct answer is $86$. Why??






limits limits-without-lhopital







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  • 2




    First multiply, tren, SUM. Is $f+gcdot h$. In your case, $6/x + 8/x *10/x = 6/x + 80/ x^2 = (6x+80)/x^2$
    – Tito Eliatron
    Nov 30 at 17:04


















1















As x approaches one, the limit of f of x is 6, the limit of g of x is
8, and the limit of h of x is 10. What is the limit as x approaches
one of the function f + g times h?




So WLOG I just set $f(x) = 6/x$, $g(x) = 8/x$ and $h(x) = 10/x$.



Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$



So $lim_{xto 1} 140/x^{2} = 140$. So my answer is $140$



But the correct answer is $86$. Why??






limits limits-without-lhopital







share|cite|improve this question


















  • 2




    First multiply, tren, SUM. Is $f+gcdot h$. In your case, $6/x + 8/x *10/x = 6/x + 80/ x^2 = (6x+80)/x^2$
    – Tito Eliatron
    Nov 30 at 17:04
















1












1








1








As x approaches one, the limit of f of x is 6, the limit of g of x is
8, and the limit of h of x is 10. What is the limit as x approaches
one of the function f + g times h?




So WLOG I just set $f(x) = 6/x$, $g(x) = 8/x$ and $h(x) = 10/x$.



Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$



So $lim_{xto 1} 140/x^{2} = 140$. So my answer is $140$



But the correct answer is $86$. Why??






limits limits-without-lhopital







share|cite|improve this question














As x approaches one, the limit of f of x is 6, the limit of g of x is
8, and the limit of h of x is 10. What is the limit as x approaches
one of the function f + g times h?




So WLOG I just set $f(x) = 6/x$, $g(x) = 8/x$ and $h(x) = 10/x$.



Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$



So $lim_{xto 1} 140/x^{2} = 140$. So my answer is $140$



But the correct answer is $86$. Why??






limits limits-without-lhopital




limits limits-without-lhopital






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 at 17:03









stackofhay42

1696




1696








  • 2




    First multiply, tren, SUM. Is $f+gcdot h$. In your case, $6/x + 8/x *10/x = 6/x + 80/ x^2 = (6x+80)/x^2$
    – Tito Eliatron
    Nov 30 at 17:04
















  • 2




    First multiply, tren, SUM. Is $f+gcdot h$. In your case, $6/x + 8/x *10/x = 6/x + 80/ x^2 = (6x+80)/x^2$
    – Tito Eliatron
    Nov 30 at 17:04










2




2




First multiply, tren, SUM. Is $f+gcdot h$. In your case, $6/x + 8/x *10/x = 6/x + 80/ x^2 = (6x+80)/x^2$
– Tito Eliatron
Nov 30 at 17:04






First multiply, tren, SUM. Is $f+gcdot h$. In your case, $6/x + 8/x *10/x = 6/x + 80/ x^2 = (6x+80)/x^2$
– Tito Eliatron
Nov 30 at 17:04












2 Answers
2






active

oldest

votes


















1















Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$




But $(f+g) cdot h ne f+g cdot h$ and they're asking:




the function f + g times h




Also: there's no need to invent functions satisfying the given limits; but if you want to, don't make it more difficult than necessary: the constant functions $6$, $8$ and $10$ do the trick!






share|cite|improve this answer





























    0














    We have that




    • $lim_{xto 1} f(x)=6 iff forall epsilon>0quad exists delta_f>0quad forall xquad 0<|x-1|<delta_f quad |f(x)-6|<epsilon$

    • $lim_{xto 1} g(x)=8iff forall epsilon>0quad exists delta_g>0quad forall xquad 0<|x-1|<delta_g quad |g(x)-8|<epsilon$

    • $lim_{xto 1} h(x)=10iff forall epsilon>0quad exists delta_h>0quad forall xquad 0<|x-1|<delta_h quad |h(x)-10|<epsilon$


    then $forall epsilon_1=3epsilon>0$ indicating by $delta=min{delta_f,delta_g,delta_h}$ we have that $forall xquad 0<|x-1|<delta$



    $$|f(x)+g(x)+h(x)-24|=|(f(x)-6)+(g(x)-8)+(h(x)-10)|le $$



    $$le|f(x)-6|+|g(x)-8|+|h(x)-10|<3epsilon=epsilon_1$$



    that is



    $$lim_{xto 1} (f(x)+g(x)+h(x))=24$$






    share|cite|improve this answer





















    • the last operation is multiplication
      – user376343
      Dec 4 at 12:25











    Your Answer





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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    1















    Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$




    But $(f+g) cdot h ne f+g cdot h$ and they're asking:




    the function f + g times h




    Also: there's no need to invent functions satisfying the given limits; but if you want to, don't make it more difficult than necessary: the constant functions $6$, $8$ and $10$ do the trick!






    share|cite|improve this answer


























      1















      Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$




      But $(f+g) cdot h ne f+g cdot h$ and they're asking:




      the function f + g times h




      Also: there's no need to invent functions satisfying the given limits; but if you want to, don't make it more difficult than necessary: the constant functions $6$, $8$ and $10$ do the trick!






      share|cite|improve this answer
























        1












        1








        1







        Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$




        But $(f+g) cdot h ne f+g cdot h$ and they're asking:




        the function f + g times h




        Also: there's no need to invent functions satisfying the given limits; but if you want to, don't make it more difficult than necessary: the constant functions $6$, $8$ and $10$ do the trick!






        share|cite|improve this answer













        Then, $(f + g) * h$ = $frac{10}{x} cdot (frac{6}{x} + frac{8}{x}) = 140/x^{2}$




        But $(f+g) cdot h ne f+g cdot h$ and they're asking:




        the function f + g times h




        Also: there's no need to invent functions satisfying the given limits; but if you want to, don't make it more difficult than necessary: the constant functions $6$, $8$ and $10$ do the trick!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 at 17:08









        StackTD

        22k1947




        22k1947























            0














            We have that




            • $lim_{xto 1} f(x)=6 iff forall epsilon>0quad exists delta_f>0quad forall xquad 0<|x-1|<delta_f quad |f(x)-6|<epsilon$

            • $lim_{xto 1} g(x)=8iff forall epsilon>0quad exists delta_g>0quad forall xquad 0<|x-1|<delta_g quad |g(x)-8|<epsilon$

            • $lim_{xto 1} h(x)=10iff forall epsilon>0quad exists delta_h>0quad forall xquad 0<|x-1|<delta_h quad |h(x)-10|<epsilon$


            then $forall epsilon_1=3epsilon>0$ indicating by $delta=min{delta_f,delta_g,delta_h}$ we have that $forall xquad 0<|x-1|<delta$



            $$|f(x)+g(x)+h(x)-24|=|(f(x)-6)+(g(x)-8)+(h(x)-10)|le $$



            $$le|f(x)-6|+|g(x)-8|+|h(x)-10|<3epsilon=epsilon_1$$



            that is



            $$lim_{xto 1} (f(x)+g(x)+h(x))=24$$






            share|cite|improve this answer





















            • the last operation is multiplication
              – user376343
              Dec 4 at 12:25
















            0














            We have that




            • $lim_{xto 1} f(x)=6 iff forall epsilon>0quad exists delta_f>0quad forall xquad 0<|x-1|<delta_f quad |f(x)-6|<epsilon$

            • $lim_{xto 1} g(x)=8iff forall epsilon>0quad exists delta_g>0quad forall xquad 0<|x-1|<delta_g quad |g(x)-8|<epsilon$

            • $lim_{xto 1} h(x)=10iff forall epsilon>0quad exists delta_h>0quad forall xquad 0<|x-1|<delta_h quad |h(x)-10|<epsilon$


            then $forall epsilon_1=3epsilon>0$ indicating by $delta=min{delta_f,delta_g,delta_h}$ we have that $forall xquad 0<|x-1|<delta$



            $$|f(x)+g(x)+h(x)-24|=|(f(x)-6)+(g(x)-8)+(h(x)-10)|le $$



            $$le|f(x)-6|+|g(x)-8|+|h(x)-10|<3epsilon=epsilon_1$$



            that is



            $$lim_{xto 1} (f(x)+g(x)+h(x))=24$$






            share|cite|improve this answer





















            • the last operation is multiplication
              – user376343
              Dec 4 at 12:25














            0












            0








            0






            We have that




            • $lim_{xto 1} f(x)=6 iff forall epsilon>0quad exists delta_f>0quad forall xquad 0<|x-1|<delta_f quad |f(x)-6|<epsilon$

            • $lim_{xto 1} g(x)=8iff forall epsilon>0quad exists delta_g>0quad forall xquad 0<|x-1|<delta_g quad |g(x)-8|<epsilon$

            • $lim_{xto 1} h(x)=10iff forall epsilon>0quad exists delta_h>0quad forall xquad 0<|x-1|<delta_h quad |h(x)-10|<epsilon$


            then $forall epsilon_1=3epsilon>0$ indicating by $delta=min{delta_f,delta_g,delta_h}$ we have that $forall xquad 0<|x-1|<delta$



            $$|f(x)+g(x)+h(x)-24|=|(f(x)-6)+(g(x)-8)+(h(x)-10)|le $$



            $$le|f(x)-6|+|g(x)-8|+|h(x)-10|<3epsilon=epsilon_1$$



            that is



            $$lim_{xto 1} (f(x)+g(x)+h(x))=24$$






            share|cite|improve this answer












            We have that




            • $lim_{xto 1} f(x)=6 iff forall epsilon>0quad exists delta_f>0quad forall xquad 0<|x-1|<delta_f quad |f(x)-6|<epsilon$

            • $lim_{xto 1} g(x)=8iff forall epsilon>0quad exists delta_g>0quad forall xquad 0<|x-1|<delta_g quad |g(x)-8|<epsilon$

            • $lim_{xto 1} h(x)=10iff forall epsilon>0quad exists delta_h>0quad forall xquad 0<|x-1|<delta_h quad |h(x)-10|<epsilon$


            then $forall epsilon_1=3epsilon>0$ indicating by $delta=min{delta_f,delta_g,delta_h}$ we have that $forall xquad 0<|x-1|<delta$



            $$|f(x)+g(x)+h(x)-24|=|(f(x)-6)+(g(x)-8)+(h(x)-10)|le $$



            $$le|f(x)-6|+|g(x)-8|+|h(x)-10|<3epsilon=epsilon_1$$



            that is



            $$lim_{xto 1} (f(x)+g(x)+h(x))=24$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 30 at 20:54









            gimusi

            1




            1












            • the last operation is multiplication
              – user376343
              Dec 4 at 12:25


















            • the last operation is multiplication
              – user376343
              Dec 4 at 12:25
















            the last operation is multiplication
            – user376343
            Dec 4 at 12:25




            the last operation is multiplication
            – user376343
            Dec 4 at 12:25


















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