Determining the limit
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Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ when $(x,y) to (2y,y)$.
I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.
calculus
edited 1 hour ago
tonychow0929
24324
asked 2 hours ago
Kashmira
373
add a comment |
up vote
2
down vote
favorite
Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ when $(x,y) to (2y,y)$.
I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.
calculus
edited 1 hour ago
tonychow0929
24324
asked 2 hours ago
Kashmira
373
1
hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago
1
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ when $(x,y) to (2y,y)$.
I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.
calculus
edited 1 hour ago
tonychow0929
24324
asked 2 hours ago
Kashmira
373
Find the limit of $$exp(frac{|x-2y|}{(x-2y)^2})$$ when $(x,y) to (2y,y)$.
I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$.
But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.
calculus
calculus
edited 1 hour ago
tonychow0929
24324
asked 2 hours ago
Kashmira
373
edited 1 hour ago
tonychow0929
24324
asked 2 hours ago
Kashmira
373
edited 1 hour ago
tonychow0929
24324
edited 1 hour ago
tonychow0929
24324
edited 1 hour ago
tonychow0929
24324
24324
asked 2 hours ago
Kashmira
373
asked 2 hours ago
Kashmira
373
asked 2 hours ago
Kashmira
373
373
1
hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago
1
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago
add a comment |
1
hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago
1
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago
1
1
hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago
hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago
1
1
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago
add a comment |
2 Answers
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4
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It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
answered 1 hour ago
Rebellos
13.8k31243
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up vote
2
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We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
answered 38 mins ago
gimusi
92.3k84495
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
4
down vote
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
answered 1 hour ago
Rebellos
13.8k31243
add a comment |
up vote
4
down vote
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
answered 1 hour ago
Rebellos
13.8k31243
add a comment |
up vote
4
down vote
up vote
4
down vote
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
answered 1 hour ago
Rebellos
13.8k31243
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
answered 1 hour ago
Rebellos
13.8k31243
answered 1 hour ago
Rebellos
13.8k31243
answered 1 hour ago
Rebellos
13.8k31243
answered 1 hour ago
Rebellos
13.8k31243
13.8k31243
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add a comment |
up vote
2
down vote
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
answered 38 mins ago
gimusi
92.3k84495
add a comment |
up vote
2
down vote
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
answered 38 mins ago
gimusi
92.3k84495
add a comment |
up vote
2
down vote
up vote
2
down vote
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
answered 38 mins ago
gimusi
92.3k84495
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
answered 38 mins ago
gimusi
92.3k84495
answered 38 mins ago
gimusi
92.3k84495
answered 38 mins ago
gimusi
92.3k84495
answered 38 mins ago
gimusi
92.3k84495
92.3k84495
add a comment |
add a comment |
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1
hint $(x-2y)^2=|x-2y|^2$
– dmtri
2 hours ago
1
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
– dmtri
2 hours ago