Multivariable Limit with ln(x)
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I need to prove that the following limit is equal to zero:
$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$
I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.
How do I go about doing this?
limits multivariable-calculus
edited Nov 17 at 7:48
Robert Z
91.5k1058129
asked Nov 17 at 7:37
Generall Josephina
31
add a comment |
up vote
0
down vote
favorite
I need to prove that the following limit is equal to zero:
$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$
I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.
How do I go about doing this?
limits multivariable-calculus
edited Nov 17 at 7:48
Robert Z
91.5k1058129
asked Nov 17 at 7:37
Generall Josephina
31
Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to prove that the following limit is equal to zero:
$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$
I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.
How do I go about doing this?
limits multivariable-calculus
edited Nov 17 at 7:48
Robert Z
91.5k1058129
asked Nov 17 at 7:37
Generall Josephina
31
I need to prove that the following limit is equal to zero:
$$lim limits_{x,y to (1,0)} frac {(x-1)^2ln(x)}{(x-1)^2 + y^2}$$
I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r cos θ$ inside the logarithmic term. I also tried using the Taylor series for $ln(x)$, but that did not simplify the expression whatsoever.
How do I go about doing this?
limits multivariable-calculus
limits multivariable-calculus
edited Nov 17 at 7:48
Robert Z
91.5k1058129
asked Nov 17 at 7:37
Generall Josephina
31
edited Nov 17 at 7:48
Robert Z
91.5k1058129
asked Nov 17 at 7:37
Generall Josephina
31
edited Nov 17 at 7:48
Robert Z
91.5k1058129
edited Nov 17 at 7:48
Robert Z
91.5k1058129
edited Nov 17 at 7:48
Robert Z
91.5k1058129
91.5k1058129
asked Nov 17 at 7:37
Generall Josephina
31
asked Nov 17 at 7:37
Generall Josephina
31
asked Nov 17 at 7:37
Generall Josephina
31
31
Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49
add a comment |
Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49
Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49
Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49
add a comment |
2 Answers
2
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begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
&le lim limits_{(x,y) to (1,0)} |ln(x)|\
&=0 end{align}
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
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up vote
0
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Let
- $u=x-1$
- $v=y$
then
$$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$
indeed
$$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$
$$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$
answered Nov 17 at 7:50
gimusi
90k74495
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
&le lim limits_{(x,y) to (1,0)} |ln(x)|\
&=0 end{align}
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
add a comment |
up vote
0
down vote
accepted
begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
&le lim limits_{(x,y) to (1,0)} |ln(x)|\
&=0 end{align}
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
&le lim limits_{(x,y) to (1,0)} |ln(x)|\
&=0 end{align}
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
begin{align}lim limits_{(x,y) to (1,0)} left|frac {(x-1)^2ln(x)}{(x-1)^2 + y^2} right|
&le lim limits_{(x,y) to (1,0)} |ln(x)|\
&=0 end{align}
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
answered Nov 17 at 7:45
Siong Thye Goh
96.1k1462116
96.1k1462116
add a comment |
add a comment |
up vote
0
down vote
Let
- $u=x-1$
- $v=y$
then
$$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$
indeed
$$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$
$$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$
answered Nov 17 at 7:50
gimusi
90k74495
add a comment |
up vote
0
down vote
Let
- $u=x-1$
- $v=y$
then
$$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$
indeed
$$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$
$$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$
answered Nov 17 at 7:50
gimusi
90k74495
add a comment |
up vote
0
down vote
up vote
0
down vote
Let
- $u=x-1$
- $v=y$
then
$$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$
indeed
$$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$
$$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$
answered Nov 17 at 7:50
gimusi
90k74495
Let
- $u=x-1$
- $v=y$
then
$$lim_{(x,y) to (1,0)} frac {(x-1)^2 ln(x)}{(x-1)^2 + y^2}=lim_{(u,v) to (0,0)} frac {u^2 ln(1+u)}{u^2 + v^2}=0$$
indeed
$$frac {u^2 ln(1+u)}{u^2 + v^2}=frac{ln(1+u)}{u}frac {u^3 }{u^2 + v^2} to 1cdot 0=0$$
$$frac {u^3 }{u^2 + v^2}=rcdot cos^3 theta to 0$$
answered Nov 17 at 7:50
gimusi
90k74495
answered Nov 17 at 7:50
gimusi
90k74495
answered Nov 17 at 7:50
gimusi
90k74495
answered Nov 17 at 7:50
gimusi
90k74495
90k74495
add a comment |
add a comment |
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Hint. $0leq frac {(x-1)^2}{(x-1)^2 + y^2}leq 1$.
– Robert Z
Nov 17 at 7:49