Determining $lim_{(x, y) to (2y, y)} exp(frac{|x-2y|}{(x-2y)^2})$
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Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ 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
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add a comment |
$begingroup$
Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ 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
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1
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hint $(x-2y)^2=|x-2y|^2$
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– dmtri
Dec 15 '18 at 8:36
1
$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39
add a comment |
$begingroup$
Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ 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
$endgroup$
Find the limit of $$expleft(frac{|x-2y|}{(x-2y)^2}right)$$ 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 Dec 15 '18 at 14:27
amWhy
1
1
asked Dec 15 '18 at 8:24
KashmiraKashmira
463
463
1
$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36
1
$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39
add a comment |
1
$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36
1
$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39
1
1
$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36
$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36
1
1
$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39
$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
$endgroup$
add a comment |
$begingroup$
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
$endgroup$
add a comment |
$begingroup$
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
$endgroup$
add a comment |
$begingroup$
It is :
$$lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{(x-2y)^2}}right) = lim_{(x,y) to (2y,y)} expleft({frac{|x-2y|}{|x-2y|^2}}right)$$
$$=$$
$$lim_{(x,y) to (2y,y)} expleft({frac{1}{|x-2y|}}right) equiv lim_{z to 0} expleft(frac{1}{|z|} right) = infty$$
$endgroup$
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 Dec 15 '18 at 8:53
RebellosRebellos
14.7k31248
14.7k31248
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$begingroup$
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
$endgroup$
add a comment |
$begingroup$
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
$endgroup$
add a comment |
$begingroup$
We have that by $t=x-2y to 0$ we reduce to the simpler
$$large e^{frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}to e^infty=infty$$
$endgroup$
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 Dec 15 '18 at 10:05
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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1
$begingroup$
hint $(x-2y)^2=|x-2y|^2$
$endgroup$
– dmtri
Dec 15 '18 at 8:36
1
$begingroup$
$e^{frac{1}{0}}=infty$ if we go frrom the right of $0$
$endgroup$
– dmtri
Dec 15 '18 at 8:39