lim(x,y)→(0,0)(1+x^2y^2)^1/x+y
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亲亲，非常荣幸为您解答
We can use polar coordinates to evaluate this limit. Let x = r cosθ and y = r sinθ. Then the limit becomes:lim(r→0) (1/(r^2 cos^2θ sin^2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/(r^2 sin2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/r)^(2/(r cosθ sinθ))= lim(r→0) e^(2ln(1/r)/(r cosθ sinθ))Now, we can use L'Hopital's rule to evaluate the exponent:lim(r→0) 2/(r^2 cosθ sinθ)= 2 lim(r→0) 1/(r^2 cosθ sinθ)= 2 lim(r→0) secθ cscθ/r^2= 0
We can use polar coordinates to evaluate this limit. Let x = r cosθ and y = r sinθ. Then the limit becomes:lim(r→0) (1/(r^2 cos^2θ sin^2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/(r^2 sin2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/r)^(2/(r cosθ sinθ))= lim(r→0) e^(2ln(1/r)/(r cosθ sinθ))Now, we can use L'Hopital's rule to evaluate the exponent:lim(r→0) 2/(r^2 cosθ sinθ)= 2 lim(r→0) 1/(r^2 cosθ sinθ)= 2 lim(r→0) secθ cscθ/r^2= 0
咨询记录 · 回答于2023-04-25
lim(x,y)→(0,0)(1+x^2y^2)^1/x+y
直线x+y+3z=0，x-y-z=0与平面x-y+z+1=0的夹角为
亲亲，非常荣幸为您解答We can use polar coordinates to evaluate this limit. Let x = r cosθ and y = r sinθ. Then the limit becomes:lim(r→0) (1/(r^2 cos^2θ sin^2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/(r^2 sin2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/r)^(2/(r cosθ sinθ))= lim(r→0) e^(2ln(1/r)/(r cosθ sinθ))Now, we can use L'Hopital's rule to evaluate the exponent:lim(r→0) 2/(r^2 cosθ sinθ)= 2 lim(r→0) 1/(r^2 cosθ sinθ)= 2 lim(r→0) secθ cscθ/r^2= 0
We can use polar coordinates to evaluate this limit. Let x = r cosθ and y = r sinθ. Then the limit becomes:lim(r→0) (1/(r^2 cos^2θ sin^2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/(r^2 sin2θ))^(1/(r cosθ sinθ))= lim(r→0) (1/r)^(2/(r cosθ sinθ))= lim(r→0) e^(2ln(1/r)/(r cosθ sinθ))Now, we can use L'Hopital's rule to evaluate the exponent:lim(r→0) 2/(r^2 cosθ sinθ)= 2 lim(r→0) 1/(r^2 cosθ sinθ)= 2 lim(r→0) secθ cscθ/r^2= 0
Therefore, the exponent approaches 0 as r approaches 0, and the limit becomes:lim(r→0) e^0= 1So the limit of the function as (x,y) approaches (0,0) is 1.
相关拓展：This limit can also be evaluated using the squeeze theorem. We can rewrite the function as:(1/(x^2y^2))^(1/(xy)) = e^(-ln(x^2y^2)/(xy))= e^(-2ln|x|/|x||y| - 2ln|y|/|x||y|)= e^(-2ln|x|/|x||y|) * e^(-2ln|y|/|x||y|)Now, we can use the inequality |ab| ≤ (a^2 + b^2)/2 to obtain:|xy| ≤ (x^2 + y^2)/2Therefore, we have:|xy|/|x||y| ≤ (x^2 + y^2)/(2|x||y|)= (x/2|y|)^2 + (y/2|x|)^2Now, as (x,y) approaches (0,0), both x/2|y| and y/2|x| approach 0. Therefore, by the squeeze theorem, the
Now, as (x,y) approaches (0,0), both x/2|y| and y/2|x| approach 0. Therefore, by the squeeze theorem, the limit of the function is also
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这边看不了图片噢
直线x+y+3z=0，x-y-z=0与平面x-y+z+1=0的夹角为
方程y^2+z^2-4x=0表示的曲面为
亲亲，非常荣幸为您解答To find the angle between the line and the plane, we first need to find the direction vector of the line and the normal vector of the plane.The line can be written in vector form as:r = twhere t is a parameter. So the direction vector of the line is:v = The plane can be written in the form:x - y + z = 1
To find the angle between the line and the plane, we first need to find the direction vector of the line and the normal vector of the plane.The line can be written in vector form as:r = twhere t is a parameter. So the direction vector of the line is:v = The plane can be written in the form:x - y + z = 1
which can also be written as: · = 1where · denotes the dot product. So the normal vector of the plane is:n = To find the angle between the line and the plane, we can use the formula:cosθ = |v · n| / (|v| |n|)where · denotes the dot product and | | denotes the magnitude.Substituting the values we found, we get:cosθ = | · | / (|| ||)= |1 - 1 + 3| / (sqrt(11) sqrt(3))= sqrt(3/11)So the angle between the line and the plan
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