evaluating limits with 2 variables
up vote
1
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Evaluate the limit :
$lim_{(x,y)to (2,2)}$ $frac{x^2 + y^2 - 8}{sqrt{x^2 +y^2} - sqrt{8}}$
$−1$
$infty$
$0$
$1$
none of the other choices
does not exist
How is the answer number 5, I thought it should be "does not exist"?
limits multivariable-calculus
edited Nov 20 at 21:02
dougle
154
asked Nov 20 at 20:19
xyz
174
add a comment |
up vote
1
down vote
favorite
Evaluate the limit :
$lim_{(x,y)to (2,2)}$ $frac{x^2 + y^2 - 8}{sqrt{x^2 +y^2} - sqrt{8}}$
$−1$
$infty$
$0$
$1$
none of the other choices
does not exist
How is the answer number 5, I thought it should be "does not exist"?
limits multivariable-calculus
edited Nov 20 at 21:02
dougle
154
asked Nov 20 at 20:19
xyz
174
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Evaluate the limit :
$lim_{(x,y)to (2,2)}$ $frac{x^2 + y^2 - 8}{sqrt{x^2 +y^2} - sqrt{8}}$
$−1$
$infty$
$0$
$1$
none of the other choices
does not exist
How is the answer number 5, I thought it should be "does not exist"?
limits multivariable-calculus
edited Nov 20 at 21:02
dougle
154
asked Nov 20 at 20:19
xyz
174
Evaluate the limit :
$lim_{(x,y)to (2,2)}$ $frac{x^2 + y^2 - 8}{sqrt{x^2 +y^2} - sqrt{8}}$
$−1$
$infty$
$0$
$1$
none of the other choices
does not exist
How is the answer number 5, I thought it should be "does not exist"?
limits multivariable-calculus
limits multivariable-calculus
edited Nov 20 at 21:02
dougle
154
asked Nov 20 at 20:19
xyz
174
edited Nov 20 at 21:02
dougle
154
asked Nov 20 at 20:19
xyz
174
edited Nov 20 at 21:02
dougle
154
edited Nov 20 at 21:02
dougle
154
edited Nov 20 at 21:02
dougle
154
154
asked Nov 20 at 20:19
xyz
174
asked Nov 20 at 20:19
xyz
174
asked Nov 20 at 20:19
xyz
174
174
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
hint
$$x^2+y^2-8=$$
$$(sqrt{x^2+y^2}+sqrt{8})(sqrt{x^2+y^2}-sqrt{8})$$
You will find $$2sqrt{8}.$$
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
add a comment |
up vote
1
down vote
HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = rcostheta$, $y = rsintheta$, and hence $r^2 = x^2 + y^2$. So, $$require{cancel}{x^2 + y^2 - 8 over sqrt{x^2 + y^2} - sqrt{8}} = {r^2 - 8over r - sqrt8} = {cancel{(r-sqrt{8})}(r + sqrt{8})over cancel{r-sqrt8}} = r+sqrt{8}.$$
answered Nov 20 at 20:34
Decaf-Math
3,104825
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
hint
$$x^2+y^2-8=$$
$$(sqrt{x^2+y^2}+sqrt{8})(sqrt{x^2+y^2}-sqrt{8})$$
You will find $$2sqrt{8}.$$
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
add a comment |
up vote
1
down vote
accepted
hint
$$x^2+y^2-8=$$
$$(sqrt{x^2+y^2}+sqrt{8})(sqrt{x^2+y^2}-sqrt{8})$$
You will find $$2sqrt{8}.$$
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
hint
$$x^2+y^2-8=$$
$$(sqrt{x^2+y^2}+sqrt{8})(sqrt{x^2+y^2}-sqrt{8})$$
You will find $$2sqrt{8}.$$
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
hint
$$x^2+y^2-8=$$
$$(sqrt{x^2+y^2}+sqrt{8})(sqrt{x^2+y^2}-sqrt{8})$$
You will find $$2sqrt{8}.$$
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
answered Nov 20 at 20:22
hamam_Abdallah
37.3k21634
37.3k21634
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
add a comment |
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
Did you rewrite the numerator? What I did was rationalized the numerator.
– xyz
Nov 20 at 20:24
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
@xyz You can simplify.
– hamam_Abdallah
Nov 20 at 20:25
add a comment |
up vote
1
down vote
HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = rcostheta$, $y = rsintheta$, and hence $r^2 = x^2 + y^2$. So, $$require{cancel}{x^2 + y^2 - 8 over sqrt{x^2 + y^2} - sqrt{8}} = {r^2 - 8over r - sqrt8} = {cancel{(r-sqrt{8})}(r + sqrt{8})over cancel{r-sqrt8}} = r+sqrt{8}.$$
answered Nov 20 at 20:34
Decaf-Math
3,104825
add a comment |
up vote
1
down vote
HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = rcostheta$, $y = rsintheta$, and hence $r^2 = x^2 + y^2$. So, $$require{cancel}{x^2 + y^2 - 8 over sqrt{x^2 + y^2} - sqrt{8}} = {r^2 - 8over r - sqrt8} = {cancel{(r-sqrt{8})}(r + sqrt{8})over cancel{r-sqrt8}} = r+sqrt{8}.$$
answered Nov 20 at 20:34
Decaf-Math
3,104825
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = rcostheta$, $y = rsintheta$, and hence $r^2 = x^2 + y^2$. So, $$require{cancel}{x^2 + y^2 - 8 over sqrt{x^2 + y^2} - sqrt{8}} = {r^2 - 8over r - sqrt8} = {cancel{(r-sqrt{8})}(r + sqrt{8})over cancel{r-sqrt8}} = r+sqrt{8}.$$
answered Nov 20 at 20:34
Decaf-Math
3,104825
HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = rcostheta$, $y = rsintheta$, and hence $r^2 = x^2 + y^2$. So, $$require{cancel}{x^2 + y^2 - 8 over sqrt{x^2 + y^2} - sqrt{8}} = {r^2 - 8over r - sqrt8} = {cancel{(r-sqrt{8})}(r + sqrt{8})over cancel{r-sqrt8}} = r+sqrt{8}.$$
answered Nov 20 at 20:34
Decaf-Math
3,104825
edited Nov 20 at 20:43
answered Nov 20 at 20:34
Decaf-Math
3,104825
answered Nov 20 at 20:34
Decaf-Math
3,104825
answered Nov 20 at 20:34
Decaf-Math
3,104825
3,104825
add a comment |
add a comment |
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