Question #74644

state whether the following statements are true or false. justify your answers.
1) lim x to 0 (1/x^2 -1/sin^2x)is in (0/0) form.
2) f(x,y)={sin(x^2y/x^3 +y^3)}/ln(x+y/x) is a homogeneous funtion of degree 2.
3) Domain of f(x,y)=xy/(x^4 +y^4) is R^2.
4) The funtion f(x,y)=(x^3y +1, x^2 +y^2) is locally invertible at (1,2).
5) The funtion f(x,y)=x^3 +y^3 is integrable on [1,2]*[1,3].

Expert's answer

Answer on Question #74644 – Math – Calculus

Question

State whether the following statements are true or false. Justify your answers.

1) lim⁡x\lim x to 0 (1/x2−1/sin⁡2x1/x^2 - 1/\sin^2 x) is in (0/0)(0/0) form.

2) f(x,y)={sin⁡(x2y/x3+y3)}/ln⁡(x+y/x)f(x,y) = \{\sin(x^2 y / x^3 + y^3)\} / \ln(x + y/x) is a homogeneous function of degree 2.

3) Domain of f(x,y)=xy/(x4+y4)f(x,y) = xy / (x^4 + y^4) is R2R^2.

4) The function f(x,y)=(x3y+1,x2y2)f(x,y) = (x^3 y + 1, x^2 y^2) is locally invertible at (1,2)(1,2).

5) The function f(x,y)=x3+y3f(x,y) = x^3 + y^3 is integrable on [1,2]×[1,3][1,2] \times [1,3].

Solution

1) lim⁡x→0(1x2−1sin⁡2x)\lim_{x\to 0}\left(\frac{1}{x^2} -\frac{1}{\sin^2x}\right) is in (00)\left(\frac{0}{0}\right) form. It is false.


lim⁡x→0(1x2−1sin⁡2x) is in (∞−∞) form butlim⁡x→0(1x2−1sin⁡2x)=lim⁡x→0(sin⁡2x−x2x2sin⁡2x)=(00)=[we can use l′Hospitale rule]=lim⁡x→0(sin⁡2x−x2)′(x2sin⁡2x)′==lim⁡x→02sin⁡(2x)−2x2sin⁡(2x)⋅x2+2x⋅sin⁡2x=(00)=[we can use l′Hospitale rule]=lim⁡x→04cos⁡(2x)−24cos⁡(2x)⋅x2+2sin⁡(2x)⋅2x+2sin⁡2x+2x⋅2sin⁡(2x)=20=∞\begin{array}{l} \lim _ {x \rightarrow 0} \left(\frac {1}{x ^ {2}} - \frac {1}{\sin^ {2} x}\right) \text{ is in } (\infty - \infty) \text{ form but} \\ \lim _ {x \rightarrow 0} \left(\frac {1}{x ^ {2}} - \frac {1}{\sin^ {2} x}\right) = \lim _ {x \rightarrow 0} \left(\frac {\sin^ {2} x - x ^ {2}}{x ^ {2} \sin^ {2} x}\right) = \left(\frac {0}{0}\right) \\ = [\text{we can use } l' \text{Hospitale rule}] = \lim _ {x \rightarrow 0} \frac {\left(\sin^ {2} x - x ^ {2}\right) ^ {\prime}}{\left(x ^ {2} \sin^ {2} x\right) ^ {\prime}} = \\ = \lim _ {x \rightarrow 0} \frac {2 \sin (2 x) - 2 x}{2 \sin (2 x) \cdot x ^ {2} + 2 x \cdot \sin^ {2} x} = \left(\frac {0}{0}\right) = [\text{we can use } l' \text{Hospitale rule}] \\ = \lim _ {x \rightarrow 0} \frac {4 \cos (2 x) - 2}{4 \cos (2 x) \cdot x ^ {2} + 2 \sin (2 x) \cdot 2 x + 2 \sin^ {2} x + 2 x \cdot 2 \sin (2 x)} = \frac {2}{0} = \infty \\ \end{array}


2) f(x,y)=sin⁡(x2yx3+y3)ln⁡(x+yx)f(x,y) = \frac{\sin\left(\frac{x^2y}{x^3 + y^3}\right)}{\ln\left(x + \frac{y}{x}\right)} is a homogeneous function of degree 2. It is false.


f(αx,αy)=sin⁡((αx)2αy(αx)3+(αy)3)ln⁡(αx+αyαx)=sin⁡(α3x2y(x3+y3)α3)ln⁡(αx+yx)==sin⁡(x2yx3+y3)ln⁡(αx+yx)≠α2f(x,y)\begin{array}{l} f (\alpha x, \alpha y) = \frac {\sin \left(\frac {(\alpha x) ^ {2} \alpha y}{(\alpha x) ^ {3} + (\alpha y) ^ {3}}\right)}{\ln \left(\alpha x + \frac {\alpha y}{\alpha x}\right)} = \frac {\sin \left(\frac {\alpha^ {3} x ^ {2} y}{(x ^ {3} + y ^ {3}) \alpha^ {3}}\right)}{\ln \left(\alpha x + \frac {y}{x}\right)} = \\ = \frac {\sin \left(\frac {x ^ {2} y}{x ^ {3} + y ^ {3}}\right)}{\ln \left(\alpha x + \frac {y}{x}\right)} \neq \alpha^ {2} f (x, y) \\ \end{array}


3) Domain of f(x,y)=xyx4+y4f(x,y) = \frac{xy}{x^4 + y^4} is R2R^2. It is false. x4+y4≠0x^4 + y^4 \neq 0. Domain of f(x,y)f(x,y) is R2/(0;0)R^2 / (0;0).

4) The function f(x,y)=[x3y+1x2+y2]f(x,y) = \begin{bmatrix} x^3 y + 1 \\ x^2 + y^2 \end{bmatrix} is locally invertible at (1,2)(1,2). This is true.

Jacobian matrix of fJf(x,y)=[∂f∂x∂f∂y∂f∂x∂f∂y]=[3x2yx32x2y]f J_{f(x,y)} = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} 3x^2 y & x^3 \\ 2x & 2y \end{bmatrix}

det(Jf(x,y))=∣3x2yx32x2y∣=6x2y2−2x4det \left(J _ {f (x, y)}\right) = \left| \begin{array}{cc} 3 x ^ {2} y & x ^ {3} \\ 2 x & 2 y \end{array} \right| = 6 x ^ {2} y ^ {2} - 2 x ^ {4}det⁡(Jf(x,y))∣y=2x=1=24−2=22≠0\left. \det \left(J _ {f (x, y)}\right) \right| _ {y = 2} ^ {x = 1} = 24 - 2 = 22 \neq 0


5) The function f(x,y)=x3+y3f(x,y) = x^{3} + y^{3} is integrable on [1;2]×[1;3][1;2] \times [1;3]. Of course this is true because x3x^{3} is continuous function on R\mathbb{R}, y3y^{3} is continuous function on R\mathbb{R}, so f(x,y)f(x,y) is continuous function on R2R^{2} and that's why it is integrable on [1;2]×[1;3][1;2] \times [1;3].

Answer:

1) false;

2) false;

3) false;

4) true;

5) true.

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