Show that the limit $lim_{(x,y)to(0,0)}frac{y+sin3x-3x}{y+x^5}$ exist/does not exist.












6














How do I show that this limit exist/does not exist? My assumption is that it does not; however I don't see how I can apply the two-path test to verify this.



$$lim_{(x,y)to(0,0)}dfrac{y+sin3x-3x}{y+x^5} = L,$$



$$lim_{yto0}dfrac{y+sin(0)-(0)}{y+(0)^5} = 1,$$



But I don't see any second path that simplifies this well;



$$lim_{xto0}dfrac{(0)+sin x-x}{(0)+x^5} = :?$$










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  • 1




    The denominator can be zero for points arbitrarily near to $(0,0)$.
    – Martín-Blas Pérez Pinilla
    Jan 29 '16 at 10:27
















6














How do I show that this limit exist/does not exist? My assumption is that it does not; however I don't see how I can apply the two-path test to verify this.



$$lim_{(x,y)to(0,0)}dfrac{y+sin3x-3x}{y+x^5} = L,$$



$$lim_{yto0}dfrac{y+sin(0)-(0)}{y+(0)^5} = 1,$$



But I don't see any second path that simplifies this well;



$$lim_{xto0}dfrac{(0)+sin x-x}{(0)+x^5} = :?$$










share|cite|improve this question




















  • 1




    The denominator can be zero for points arbitrarily near to $(0,0)$.
    – Martín-Blas Pérez Pinilla
    Jan 29 '16 at 10:27














6












6








6


2





How do I show that this limit exist/does not exist? My assumption is that it does not; however I don't see how I can apply the two-path test to verify this.



$$lim_{(x,y)to(0,0)}dfrac{y+sin3x-3x}{y+x^5} = L,$$



$$lim_{yto0}dfrac{y+sin(0)-(0)}{y+(0)^5} = 1,$$



But I don't see any second path that simplifies this well;



$$lim_{xto0}dfrac{(0)+sin x-x}{(0)+x^5} = :?$$










share|cite|improve this question















How do I show that this limit exist/does not exist? My assumption is that it does not; however I don't see how I can apply the two-path test to verify this.



$$lim_{(x,y)to(0,0)}dfrac{y+sin3x-3x}{y+x^5} = L,$$



$$lim_{yto0}dfrac{y+sin(0)-(0)}{y+(0)^5} = 1,$$



But I don't see any second path that simplifies this well;



$$lim_{xto0}dfrac{(0)+sin x-x}{(0)+x^5} = :?$$







calculus real-analysis limits






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edited Nov 28 '18 at 12:30









Lorenzo B.

1,8322520




1,8322520










asked Jan 29 '16 at 10:17









Frank Vel

2,34942246




2,34942246








  • 1




    The denominator can be zero for points arbitrarily near to $(0,0)$.
    – Martín-Blas Pérez Pinilla
    Jan 29 '16 at 10:27














  • 1




    The denominator can be zero for points arbitrarily near to $(0,0)$.
    – Martín-Blas Pérez Pinilla
    Jan 29 '16 at 10:27








1




1




The denominator can be zero for points arbitrarily near to $(0,0)$.
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27




The denominator can be zero for points arbitrarily near to $(0,0)$.
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27










1 Answer
1






active

oldest

votes


















2














Hint



Let $f(x,y)=frac{y+sin(3x)-3x}{y+x^5}$.



What do you think about $$lim_{tto 0}f(0,t)quad text{and}quad lim_{tto 0}f(t,0) ?$$



Edited



$$sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$$
therefore $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$






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  • Am I allowed to use L'Hôpital on the second one?
    – Frank Vel
    Jan 29 '16 at 10:22






  • 1




    It's indeed a possibility.
    – Surb
    Jan 29 '16 at 10:23











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Hint



Let $f(x,y)=frac{y+sin(3x)-3x}{y+x^5}$.



What do you think about $$lim_{tto 0}f(0,t)quad text{and}quad lim_{tto 0}f(t,0) ?$$



Edited



$$sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$$
therefore $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$






share|cite|improve this answer























  • Am I allowed to use L'Hôpital on the second one?
    – Frank Vel
    Jan 29 '16 at 10:22






  • 1




    It's indeed a possibility.
    – Surb
    Jan 29 '16 at 10:23
















2














Hint



Let $f(x,y)=frac{y+sin(3x)-3x}{y+x^5}$.



What do you think about $$lim_{tto 0}f(0,t)quad text{and}quad lim_{tto 0}f(t,0) ?$$



Edited



$$sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$$
therefore $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$






share|cite|improve this answer























  • Am I allowed to use L'Hôpital on the second one?
    – Frank Vel
    Jan 29 '16 at 10:22






  • 1




    It's indeed a possibility.
    – Surb
    Jan 29 '16 at 10:23














2












2








2






Hint



Let $f(x,y)=frac{y+sin(3x)-3x}{y+x^5}$.



What do you think about $$lim_{tto 0}f(0,t)quad text{and}quad lim_{tto 0}f(t,0) ?$$



Edited



$$sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$$
therefore $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$






share|cite|improve this answer














Hint



Let $f(x,y)=frac{y+sin(3x)-3x}{y+x^5}$.



What do you think about $$lim_{tto 0}f(0,t)quad text{and}quad lim_{tto 0}f(t,0) ?$$



Edited



$$sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$$
therefore $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 29 '16 at 10:23

























answered Jan 29 '16 at 10:19









Surb

37.4k94375




37.4k94375












  • Am I allowed to use L'Hôpital on the second one?
    – Frank Vel
    Jan 29 '16 at 10:22






  • 1




    It's indeed a possibility.
    – Surb
    Jan 29 '16 at 10:23


















  • Am I allowed to use L'Hôpital on the second one?
    – Frank Vel
    Jan 29 '16 at 10:22






  • 1




    It's indeed a possibility.
    – Surb
    Jan 29 '16 at 10:23
















Am I allowed to use L'Hôpital on the second one?
– Frank Vel
Jan 29 '16 at 10:22




Am I allowed to use L'Hôpital on the second one?
– Frank Vel
Jan 29 '16 at 10:22




1




1




It's indeed a possibility.
– Surb
Jan 29 '16 at 10:23




It's indeed a possibility.
– Surb
Jan 29 '16 at 10:23


















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