Show that the limit $lim_{(x,y)to(0,0)}frac{y+sin3x-3x}{y+x^5}$ exist/does not exist.
$begingroup$
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
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
$endgroup$
add a comment |
$begingroup$
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
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
$endgroup$
1
$begingroup$
The denominator can be zero for points arbitrarily near to $(0,0)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27
add a comment |
$begingroup$
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
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
$endgroup$
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
calculus real-analysis limits
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
edited Nov 28 '18 at 12:30
Lorenzo B.
1,8402520
1,8402520
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
asked Jan 29 '16 at 10:17
Frank VelFrank Vel
2,34942246
2,34942246
1
$begingroup$
The denominator can be zero for points arbitrarily near to $(0,0)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27
add a comment |
1
$begingroup$
The denominator can be zero for points arbitrarily near to $(0,0)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27
1
1
$begingroup$
The denominator can be zero for points arbitrarily near to $(0,0)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27
$begingroup$
The denominator can be zero for points arbitrarily near to $(0,0)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
$endgroup$
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
1
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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$begingroup$
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 $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
$endgroup$
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
1
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
add a comment |
$begingroup$
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 $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
$endgroup$
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
1
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
add a comment |
$begingroup$
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 $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
$endgroup$
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 $$frac{sin(x)-x}{x^5}=frac{-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)}{x^5}underset{xto 0}{longrightarrow }-infty .$$
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
edited Jan 29 '16 at 10:23
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
answered Jan 29 '16 at 10:19
SurbSurb
37.7k94375
37.7k94375
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
1
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
add a comment |
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
1
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
$begingroup$
Am I allowed to use L'Hôpital on the second one?
$endgroup$
– Frank Vel
Jan 29 '16 at 10:22
1
1
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
$begingroup$
It's indeed a possibility.
$endgroup$
– Surb
Jan 29 '16 at 10:23
add a comment |
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$begingroup$
The denominator can be zero for points arbitrarily near to $(0,0)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Jan 29 '16 at 10:27