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Evaluating the limit for a point on the curve
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For point p(a,b) lying on the curve $2xy^2dx + 2x^2 y dy -
tan(x^2y^2) dx =0 $;
$lim_ato -inftyb =? $
Options are:
a) $ 0$
b) $-1 $
c) $1$
d) does not exist.
Attempt:
If we observe carefully we get:
$d(x^2 y^2) = tan(x^2 y^2) dx$
$implies ln(csin x^2y^2) = x$
$implies c sin (x^2 y^2) = e^x$
Now clearly as $x to -infty ~ , e^x to 0$, so clearly $y to 0$
But answer given is d. Please let me know my mistake.
calculus
asked 37 mins ago
Abcd
2,6691926
add a comment |Â
up vote
5
down vote
favorite
For point p(a,b) lying on the curve $2xy^2dx + 2x^2 y dy -
tan(x^2y^2) dx =0 $;
$lim_ato -inftyb =? $
Options are:
a) $ 0$
b) $-1 $
c) $1$
d) does not exist.
Attempt:
If we observe carefully we get:
$d(x^2 y^2) = tan(x^2 y^2) dx$
$implies ln(csin x^2y^2) = x$
$implies c sin (x^2 y^2) = e^x$
Now clearly as $x to -infty ~ , e^x to 0$, so clearly $y to 0$
But answer given is d. Please let me know my mistake.
calculus
asked 37 mins ago
Abcd
2,6691926
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
For point p(a,b) lying on the curve $2xy^2dx + 2x^2 y dy -
tan(x^2y^2) dx =0 $;
$lim_ato -inftyb =? $
Options are:
a) $ 0$
b) $-1 $
c) $1$
d) does not exist.
Attempt:
If we observe carefully we get:
$d(x^2 y^2) = tan(x^2 y^2) dx$
$implies ln(csin x^2y^2) = x$
$implies c sin (x^2 y^2) = e^x$
Now clearly as $x to -infty ~ , e^x to 0$, so clearly $y to 0$
But answer given is d. Please let me know my mistake.
calculus
asked 37 mins ago
Abcd
2,6691926
For point p(a,b) lying on the curve $2xy^2dx + 2x^2 y dy -
tan(x^2y^2) dx =0 $;
$lim_ato -inftyb =? $
Options are:
a) $ 0$
b) $-1 $
c) $1$
d) does not exist.
Attempt:
If we observe carefully we get:
$d(x^2 y^2) = tan(x^2 y^2) dx$
$implies ln(csin x^2y^2) = x$
$implies c sin (x^2 y^2) = e^x$
Now clearly as $x to -infty ~ , e^x to 0$, so clearly $y to 0$
But answer given is d. Please let me know my mistake.
calculus
calculus
asked 37 mins ago
Abcd
2,6691926
asked 37 mins ago
Abcd
2,6691926
asked 37 mins ago
Abcd
2,6691926
asked 37 mins ago
Abcd
2,6691926
asked 37 mins ago
Abcd
2,6691926
2,6691926
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
We have the following line:
$$c sin (x^2 y^2) = e^x$$
The problem is that $arcsin$ return only a single branch of $sin$, not all of them:
Now, $c sin (x^2 y^2) = e^x$ gives $$sin (x^2 y^2) = frace^xcimplies x^2 y^2=arcsinleft(frace^xcright)colorred+2kpi\implies y=pmfrac1xsqrtarcsinleft(frace^xcright)colorred+2kpi$$So you have more then a single value for $y$ for any fixed $x$ and the answered are as big as we want so the limit won't get rid from them. So the limit is undefined
answered 24 mins ago
Holo
4,8352829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
We have the following line:
$$c sin (x^2 y^2) = e^x$$
The problem is that $arcsin$ return only a single branch of $sin$, not all of them:
Now, $c sin (x^2 y^2) = e^x$ gives $$sin (x^2 y^2) = frace^xcimplies x^2 y^2=arcsinleft(frace^xcright)colorred+2kpi\implies y=pmfrac1xsqrtarcsinleft(frace^xcright)colorred+2kpi$$So you have more then a single value for $y$ for any fixed $x$ and the answered are as big as we want so the limit won't get rid from them. So the limit is undefined
answered 24 mins ago
Holo
4,8352829
add a comment |Â
up vote
3
down vote
We have the following line:
$$c sin (x^2 y^2) = e^x$$
The problem is that $arcsin$ return only a single branch of $sin$, not all of them:
Now, $c sin (x^2 y^2) = e^x$ gives $$sin (x^2 y^2) = frace^xcimplies x^2 y^2=arcsinleft(frace^xcright)colorred+2kpi\implies y=pmfrac1xsqrtarcsinleft(frace^xcright)colorred+2kpi$$So you have more then a single value for $y$ for any fixed $x$ and the answered are as big as we want so the limit won't get rid from them. So the limit is undefined
answered 24 mins ago
Holo
4,8352829
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have the following line:
$$c sin (x^2 y^2) = e^x$$
The problem is that $arcsin$ return only a single branch of $sin$, not all of them:
Now, $c sin (x^2 y^2) = e^x$ gives $$sin (x^2 y^2) = frace^xcimplies x^2 y^2=arcsinleft(frace^xcright)colorred+2kpi\implies y=pmfrac1xsqrtarcsinleft(frace^xcright)colorred+2kpi$$So you have more then a single value for $y$ for any fixed $x$ and the answered are as big as we want so the limit won't get rid from them. So the limit is undefined
answered 24 mins ago
Holo
4,8352829
We have the following line:
$$c sin (x^2 y^2) = e^x$$
The problem is that $arcsin$ return only a single branch of $sin$, not all of them:
Now, $c sin (x^2 y^2) = e^x$ gives $$sin (x^2 y^2) = frace^xcimplies x^2 y^2=arcsinleft(frace^xcright)colorred+2kpi\implies y=pmfrac1xsqrtarcsinleft(frace^xcright)colorred+2kpi$$So you have more then a single value for $y$ for any fixed $x$ and the answered are as big as we want so the limit won't get rid from them. So the limit is undefined
answered 24 mins ago
Holo
4,8352829
answered 24 mins ago
Holo
4,8352829
answered 24 mins ago
Holo
4,8352829
answered 24 mins ago
Holo
4,8352829
4,8352829
add a comment |Â
add a comment |Â
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